1. Introduction

In 1950, Erdös ^[1] conjectured for any positive integer $n>1$, that the following diophantine equation

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{4}{n}$

(1.1)

has positive integer solutions x, y, z. Later, Strauss ^[1] made a more powerful conjecture: let $n>2$, then diophantine equation (1.1) has positive integer solution $x, y, z$ with $x\neq y, x\neq z, y\neq z$. He proved that the conjecture is true with $n＜5000$. In 1964, Zhao Ke, Qi Sun and Xianjue Zhang ^[3] proved that Strauss conjecture is equivalent to Erdös conjecture. In 1979, Ke and Sun ^[4] proved that Erdös-Strauss conjecture is true with $n＜4\cdot 10^{5}$. In 1965, Yamamoto ^[10] proved that Erdös-Strauss conjecture is also true with $n＜10^{7}$. In 1978, Franceschine ^[2] proved that Erdös-Strauss conjecture is true with $n＜10^{8}$. Sierpiski made a similar conjecture: for any positive integer $n＞1$, that the following diophantine equation

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{n}$

(1.2)

has the solutions $x, y, z$ of positive integer. Palama ^[6,7] proved that Sierpiski conjecture is true with $n＜922321$.Stewart ^[9] also obtained above the result with $n\leq 105743881$ and $n\not\equiv1\pmod{278460}$. In 1984, Liu ^[5] obtained all the solution of positive integers of the following diophantine equation

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{121}.$

(1.3)

Write $n=12p$ in (1.1) or write $n=15p$ in (1.2), where $p$ is an odd prime. We obtain diophantine equation

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{3p}, x\le y\le z.$

(1.4)

One can easily get all the solutions of positive integer of (1.4) when $p|x$. If $3|x,p\nmid x$, then one lets

$x=3a{{x}_{1}}, y=a{{y}_{1}}, z=a{{z}_{1}}, a=\gcd (x/3, y, z),$

and so (1.4) is changed to

$3p\left( {{y}_{1}}+{{z}_{1}} \right){{x}_{1}}=\left( a{{x}_{1}}-p \right){{y}_{1}}{{z}_{1}}.$

(1.5)

Moreover, we write $y_1=dy_2, z_1=dz_2, d=\gcd(y_1, z_1)$. It is easy to see that $\gcd(d, x_1)=1$.

If $3p|d$, then we get that $x_1|y_2z_2$ and $y_2z_2|x_1$. It follows that $x_1=y_2z_2$. Write $\frac{d}{3p}=d_1$, then (1.5) is changed to

$\frac{{{y}_{2}}+{{z}_{2}}}{{{d}_{1}}}=a{{y}_{2}}{{z}_{2}}-p, \gcd ({{y}_{2}}, {{z}_{2}})=1, {{y}_{2}}\le {{z}_{2}}.$

If $3\nmid d,p|d$ and $3|y_1$, then we get that $x_1|y_3z_2$ and $y_3z_2|x_1$ where $y_2=3y_3$. It follows that $x_1=y_3z_2$. Write $\frac{d}{p}=d_1$, then (1.5) is changed to

$\frac{3{{y}_{3}}+{{z}_{2}}}{{{d}_{1}}}=a{{y}_{3}}{{z}_{2}}-p, \gcd (3{{y}_{3}}, {{z}_{2}})=1, 3{{y}_{3}}\le {{z}_{2}}.$

If $3\nmid d,p|d$ and $3|z_1$, then we get that $x_1|y_2z_3$ and $y_2z_3|x_1$ where $z_2=3z_3$. It follows that $x_1=y_2z_3$. Write $\frac{d}{p}=d_1$, then (1.5) is changed to

$\frac{{{y}_{2}}+3{{z}_{3}}}{{{d}_{1}}}=a{{y}_{2}}{{z}_{3}}-p, \gcd ({{y}_{2}}, 3{{z}_{3}})=1.$

If $p\nmid d,3|d$ and $p|y_1$, then we get that $x_1|y_3z_2$ and $y_3z_2|x_1$ where $y_2=py_3$. It follows that $x_1=y_3z_2$. Write $\frac{d}{3}=d_1$, then (1.5) is changed to

$\frac{p{{y}_{3}}+{{z}_{2}}}{{{d}_{1}}}=a{{y}_{3}}{{z}_{2}}-p, \gcd (p{{y}_{3}}, {{z}_{2}})=1, p{{y}_{3}}\le {{z}_{2}}.$

If $p\nmid d,3|d$ and $p|z_1$, then we get that $x_1|y_2z_3$ and $y_2z_3|x_1$ where $z_2=pz_3$. It follows that $x_1=y_2z_3$. Write $\frac{d}{3}=d_1$, then (1.5) is changed to

$\frac{{{y}_{2}}+p{{z}_{3}}}{{{d}_{1}}}=a{{y}_{2}}{{z}_{3}}-p, \gcd ({{y}_{2}}, p{{z}_{3}})=1.$

If $3p\nmid d$ and $3|y_1, p|z_1$, then we get that $x_1|y_3z_3$ and $y_3z_3|x_1$ where $y_2=3y_3, z_2=pz_3$. It follows that $x_1=y_3z_3$. Then (1.5) is changed to

$\frac{3{{y}_{3}}+p{{z}_{3}}}{d}=a{{y}_{3}}{{z}_{3}}-p, \gcd (3{{y}_{3}}, p{{z}_{3}})=1.$

If $3p\nmid d$ and $p|y_1, 3|z_1$, then we get that $x_1|y_3z_3$ and $y_3z_3|x_1$ where $y_2=py_3, z_2=3z_3$. It follows that $x_1=y_3z_3$. Then (1.5) is changed to

$\frac{p{{y}_{3}}+3{{z}_{3}}}{d}=a{{y}_{3}}{{z}_{3}}-p, \gcd (p{{y}_{3}}, 3{{z}_{3}})=1.$

If $3\nmid y_1z_1, p\nmid d$ and $p|y_1$, then we get that $x_1|y_3z_2$ and $y_3z_2|x_1$ where $y_2=py_3$. It follows that $x_1=y_3z_2$. Then (1.5) is changed to

$\frac{p{{y}_{3}}+{{z}_{2}}}{d}=\frac{a{{y}_{3}}{{z}_{2}}-p}{3}, \gcd (p{{y}_{3}}, {{z}_{2}})=1.$

If $3\nmid y_1z_1, p\nmid d$ and $p|z_1$, then we get that $x_1|y_2z_3$ and $y_2z_3|x_1$ where $z_2=pz_3$. It follows that $x_1=y_2z_3$. Then (1.5) is changed to

$\frac{{{y}_{2}}+p{{z}_{3}}}{d}=\frac{a{{y}_{2}}{{z}_{3}}-p}{3}, \gcd ({{y}_{2}}, p{{z}_{3}})=1.$

If $3\nmid y_1z_1, p|d$, then we get that $x_1|y_2z_2$ and $y_2z_2|x_1$. It follows that $x_1=y_2z_2$. Write $\frac{d}{p}=d_1$, then (1.5) is changed to

$\frac{{{y}_{2}}+{{z}_{2}}}{{{d}_{1}}}=\frac{a{{y}_{2}}{{z}_{2}}-p}{3}, \gcd ({{y}_{2}}, {{z}_{2}})=1.$

Thus we have proved that solving the equation (1.5) is equivalent to solving the following diophantine equations:

$\frac{y+z}{d}=ayz-p, \gcd (y, z)=1, y\le z,$

(1.6)

$\frac{3y+z}{d}=ayz-p, \gcd (3y, z)=1,$

(1.7)

$\frac{py+z}{d}=ayz-p, \gcd (py, z)=1,$

(1.8)

$\frac{3y+pz}{d}=ayz-p, \gcd (3y, pz)=1,$

(1.9)

$\frac{py+z}{d}=\frac{ayz-p}{3}, \gcd (py, z)=1, 3\nmid yz,$

(1.10)

$\frac{y+z}{d}=\frac{ayz-p}{3}, \gcd (y, z)=1, y\le z, 3\nmid yz.$

(1.11)

In this paper, we investigate the equations (1.6), (1.7) and (1.8) with $p=661$. Actually, we get all the solutions of positive integer of them. That is, we have the following results:

Theorem 1.1. If $ayz-661\geq 80$, then all the solutions of positive integers of (1.6) are given by $(a, d, y, z)=$

(1, 1, 2, 663), (1, 1, 3, 332), (1, 2, 1, 1323), (1, 2, 3, 265), (1, 3, 1, 992), (1, 3, 2, 397),

(2, 1, 1, 662), (2, 1, 2, 221), (2, 1, 4, 95), (2, 2, 1, 441), (3, 1, 1, 331),

(3, 1, 3, 83), (3, 3, 1, 248)(4, 2, 1, 189).

Theorem 1.2. If $ayz-661\geq 90$, then all the solutions of positive integers of (1.7) are given by $(a, d, y, z)=$

(1, 1, 2, 667), (1, 1, 3, 335), (1, 1, 5, 169), (1, 1, 9, 86), (1, 1, 84, 11), (1, 1, 167, 7), (1, 1, 333, 5),

(1, 1, 665, 4), (1, 2, 1, 1325), (1, 4, 2645, 1), (1, 5, 1, 827), (1, 5, 1653, 1), (1, 7, 1157, 1), (1, 11, 909, 1),

(1, 19, 785, 1), (2, 1, 1, 664), (2, 1, 3, 134), (2, 1, 27, 14), (2, 1, 133, 4), (2, 1, 663, 2), (2, 2, 1323, 1),

(2, 4, 529, 1), (3, 1, 1, 332), (3, 1, 221, 2), (3, 2, 1, 265), (3, 2, 441, 1), (4, 1, 662, 1), (5, 1, 1, 166),

(5, 1, 331, 1), (5, 2, 189, 1), (6, 2, 147, 1), (8, 1, 51, 2), (10, 1, 39, 2), (12, 2, 63, 1).

Theorem 1.3. Each of the following is true:

1. If $ayz-661＜80$, then all the solutions of positive integers of (1.6) are given by $(a, d, y, z)=$

(1, 3, 5, 142), (1, 3, 17, 40), (1, 7, 6, 113), (1, 11, 18, 37), (1, 28, 17, 39), (1, 32, 13, 51),

(1, 67, 2, 333), (1, 112, 3, 221), (1, 332, 1, 663), (1, 333, 2, 331), (1, 663, 1, 662), (2, 1, 5, 74),

(2, 1, 11, 32), (2, 1, 14, 25), (2, 2, 11, 31), (2, 7, 1, 356), (2, 8, 5, 67), (2, 20, 1, 339), (2, 26, 1, 337),

(2, 29, 4, 83), (2, 112, 1, 332), (2, 332, 1, 331), (3, 1, 11, 21), (3, 14, 13, 17), (3, 28, 1, 223),

(3, 111, 1, 221), (4, 1, 6, 29), (4, 5, 1, 174), (4, 24, 1, 167), (5, 1, 4, 35), (5, 1, 6, 23), (5, 2, 1, 147),

(5, 15, 1, 134), (6, 4, 1, 115), (6, 8, 3, 37), (7, 1, 2, 51), (7, 6, 5, 19), (7, 24, 1, 95), (8, 28, 1, 83),

(9, 1, 2, 39), (9, 1, 4, 19), (9, 2, 3, 25), (9, 15, 1, 74), (11, 2, 1, 63), (11, 3, 1, 62), (13, 10, 3, 17),

(13, 26, 1, 51), (14, 2, 1, 49), (15, 1, 5, 9), (17, 8, 3, 13), (17, 20, 1, 39), (19, 3, 5, 7), (19, 9, 1, 35),

(20, 1, 2, 17), (21, 3, 1, 32), (23, 5, 1, 29), (24, 1, 4, 7), (25, 2, 1, 27), (26, 1, 2, 13), (28, 1, 3, 8),

(29, 4, 1, 23), (32, 2, 1, 21), (35, 5, 1, 19), (39, 9, 1, 17), (51, 7, 1, 13), (74, 2, 1, 9), (83, 3, 1, 8),

(95, 2, 1, 7), (111, 1, 2, 3), (221, 2, 1, 3), (331, 3, 1, 2), (332, 1, 1, 2), (662, 2, 1, 1), (663, 1, 1, 1).

2. If $ayz-661＜90$, then all the solutions of positive integers of (1.7) are given by $(a, d, y, z)=$

(1, 9, 1, 744), (1, 29, 3, 223), (1, 31, 35, 19), (1, 37, 5, 133), (1, 67, 39, 17), (1, 73, 95, 7),

(1, 83, 51, 13), (1, 101, 133, 5), (1, 115, 3, 221), (1, 127, 677, 1), (1, 167, 1, 665), (1, 333, 1, 663),

(1, 337, 2, 331), (1, 665, 1, 662), (1, 995, 331, 2), (1, 995, 663, 1), (1, 1987, 662, 1), (2, 1, 13, 28),

(2, 2, 3, 121), (2, 200, 333, 1), (2, 334, 1, 331), (2, 994, 331, 1), (3, 2, 4, 58), (3, 7, 2, 113),

(3, 11, 6, 37), (3, 28, 13, 17), (3, 32, 17, 13), (3, 67, 111, 2), (3, 112, 1, 221), (3, 332, 221, 1),

(4, 2, 9, 19), (5, 10, 7, 19), (5, 16, 19, 7), (5, 34, 1, 133), (5, 100, 133, 1), (6, 5, 16, 7), (6, 20, 113, 1),

(8, 1, 18, 5), (8, 23, 84, 1), (9, 1, 1, 83), (9, 1, 7, 11), (12, 1, 2, 29), (12, 5, 58, 1), (13, 2, 1, 53),

(13, 13, 3, 17), (15, 1, 2, 23), (15, 2, 49, 1), (17, 11, 3, 13), (17, 59, 39, 1), (18, 8, 1, 37), (26, 2, 27, 1),

(27, 1, 13, 2), (27, 2, 1, 25), (33, 2, 21, 1), (34, 1, 5, 4), (37, 11, 18, 5), (39, 10, 1, 17), (39, 26, 17, 1),

(45, 1, 3, 5), (51, 8, 1, 13), (51, 20, 13, 1), (75, 2, 9, 1), (84, 1, 1, 8), (96, 2, 7, 1), (112, 1, 3, 2),

(133, 2, 1, 5), (167, 1, 1, 4), (221, 5, 3, 1), (222, 2, 3, 1), (331, 5, 1, 2), (331, 7, 2, 1), (333, 1, 1, 2),

(334, 1, 2, 1), (662, 4, 1, 1), (663, 2, 1, 1), (665, 1, 1, 1).

Theorem 1.4. All the solutions of positive integers of (1.8) with $py\leq z$ are given by $(a, d, y, z)=$

(1, 1, 2, 1983), (1, 2, 1, 1983), (1, 3, 1, 1322), (1, 3, 2, 661), (1, 331, 2, 333), (1, 662, 1, 663),

(1, 1323, 1, 662), (1, 1653, 2, 331), (2, 1, 1, 1322), (2, 2, 1, 661), (2, 331, 1, 332), (2, 992, 1, 331).

We organize this paper as follows. In Section 2, we present some lemmas which are needed in the proof of our main results. Consequently, in Sections 3 to 6, we give the proofs of Theorem 1.1 to 1.4 respectively.

---

2. Some lemmas

To prove the main theorems, we need the following lemmas.

Lemma 2.1. Let $661+b=p$, where $p$ is an odd prime. If one of the following conditions is satisfied

$\begin{align} & \begin{matrix} p\equiv 1 & \left( \bmod q \right), q=3, 5 or 11 or p\equiv 3 & \left( \bmod 7 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv 23 & \left( \bmod 29 \right), or p\equiv 5 & \left( \bmod 41 \right), \\ \end{matrix} \\ \end{align}$

then both equation (1.6) and equation (1.7) have no solution (a, d, y, z) of positive integers with $ayz-661=b$.

proof. We only prove the case $p\equiv23\pmod{29}$. The proofs of other cases are similarly. By the assumption we have $b\equiv p-661\equiv0\pmod{29}$. Assume either (1.6) or (1.7) has a positive integer solution $(a, d, y, z)$ with $ayz-661=b$. It follows $(y, z)\in \{(1, 1), (1, p), (p, 1)\}$ and either

$\begin{matrix} y+z\equiv 0 & \left( \bmod 29 \right) \\ \end{matrix}$

(2.1)

or

$\begin{matrix} 3y+z\equiv 0 & \left( \bmod 29 \right). \\ \end{matrix}$

(2.2)

One can easily to see neither (2.1) nor (2.2) is satisfied for any element $(y, z)\in \{(1, 1), (1, p), (p, 1)\}$. This completes the proof of Lemma 2.1}.

Lemma 2.2. Let $661+b=pq$, where $p, q$ are prime numbers.

(i)If one of the following conditions is satisfied

$\begin{align} & \begin{matrix} p\equiv q\equiv 1 & \left( \bmod 3 \right), or p\equiv q\equiv 1 & \left( \bmod 5 \right), or p\equiv 1 & \left( \bmod 7 \right), q\equiv 3 & \left( \bmod 7 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv 3 & \left( \bmod 13 \right), q\equiv -5 & \left( \bmod 13 \right), or p\equiv 3 & \left( \bmod 17 \right), q\equiv 5 & \left( \bmod 17 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv 3 & \left( \bmod 19 \right), q\equiv 5 & \left( \bmod 19 \right), or p\equiv 3 & \left( \bmod 20 \right), q\equiv 7 & \left( \bmod 20 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv 7 & \left( \bmod 23 \right), q\equiv 9 & \left( \bmod 23 \right), or p\equiv 3 & \left( \bmod 31 \right), q\equiv -7 & \left( \bmod 31 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv 2 & \left( \bmod 37 \right), q\equiv 14 & \left( \bmod 37 \right), or p\equiv 23 & \left( \bmod 52 \right), q\equiv 31 & \left( \bmod 52 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv 3 & \left( \bmod 56 \right), q\equiv 15 & \left( \bmod 56 \right), or p\equiv 2 & \left( \bmod 57 \right), q\equiv 17 & \left( \bmod 57 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv q\equiv 2 & \left( \bmod 73 \right), or p\equiv 2 & \left( \bmod 85 \right), q\equiv 33 & \left( \bmod 85 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv 7 & \left( \bmod 88 \right), q\equiv 19 & \left( \bmod 88 \right), \\ \end{matrix} \\ \end{align}$

then both equation (1.6) and equation (1.7) have no positive integer solution (a, d, y, z) with $ayz-661=b$.

(ii). If one of the following conditions is satisfied

$\begin{align} & \begin{matrix} p\equiv q\equiv 1 & \left( \bmod 4 \right), or p\equiv q\equiv \pm 2 & \left( \bmod 9 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv 2 & \left( \bmod 11 \right), q\equiv 6 & \left( \bmod 11 \right), \\ \end{matrix} \\ \end{align}$

then equation (1.6) has no positive integer solution (a, d, y, z) with $ayz-661=b$.

proof. (i) We only prove the case $p\equiv7\pmod{23}, q\equiv9\pmod{23}$.The proofs of other cases are similarly. By the assumption we have $b\equiv pq-661\equiv0\pmod{23}$. Assume either (1.6) or (1.7) has a positive integer solution $(a, d, y, z)$ with $ayz-661=b$. It follows $(y, z)\in \{(1, p^{u}q^{v}), (p, q^{v}), (q, p^{u}), (pq, 1), u, v=0, 1\}$ and either

$\begin{matrix} y+z\equiv 0 & \left( \bmod 23 \right) \\ \end{matrix}$

(2.3)

or

$3y+z\equiv 0\begin{matrix} {} & \left( \bmod 23 \right). \\ \end{matrix}$

(2.4)

One can easily to see neither (2.3) nor (2.4) is satisfied for any element $(y, z)\in \{(1, p^{u}q^{v}), (p, q^{v}), (q, p^{u}), (pq, 1), u, v=0, 1\}$. This completes the proof of part (i) of Lemma 2.2.

The proof of part (ii) is similar.

Lemma 2.3. Let $661+b=pqr$, where $p, q, r$ are prime. If one of the following conditions is satisfied

$\begin{align} & \begin{matrix} p\equiv q\equiv 3 & \left( \bmod 11 \right), r\equiv 5 & \left( \bmod 11 \right), or p\equiv q\equiv 3 & \left( \bmod 25 \right), r\equiv 4 & \left( \bmod 25 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv q\equiv 2 & \left( \bmod 31 \right), r\equiv 18 & \left( \bmod 31 \right), or p\equiv q\equiv 2 & \left( \bmod 55 \right), r\equiv 14 & \left( \bmod 55 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv q\equiv 19 & \left( \bmod 61 \right), r\equiv 2 & \left( \bmod 61 \right), or p\equiv q\equiv 5 & \left( \bmod 64 \right), r\equiv 29 & \left( \bmod 64 \right), \\ \end{matrix} \\ & \begin{matrix} or p\equiv 3 & \left( \bmod 80 \right), q\equiv 13 & \left( \bmod 80 \right), r\equiv 19 & \left( \bmod 80 \right), \\ \end{matrix} \\ \end{align}$

then both equation (1.6) and equation (1.7) have no positive integer solution (a, d, y, z) with $ayz-661=b$.

proof. We only prove the case $p\equiv q\equiv19\pmod{61}, r\equiv2\pmod{61}$. The proofs of other cases are similarly. By the assumption we have $b\equiv pqr-661\equiv0\pmod{61}$. Assume either (1.6) or (1.7) has a positive integer solution $(a, d, y, z)$ with $ayz-661=b$. It follows

$(y, z)\in \{(1, {{p}^{u}}{{q}^{v}}{{r}^{t}}), (p, {{q}^{v}}{{r}^{t}}), (q, {{p}^{u}}{{r}^{t}}), (r, {{p}^{u}}{{q}^{v}}), (pq, {{r}^{t}}), (pr, {{q}^{v}}), (qr, {{p}^{t}}), (pqr, 1), u, v, t=0, 1\}$

and either

$\begin{matrix} y+z\equiv 0 & \left( \bmod 61 \right) \\ \end{matrix}$

(2.5)

or

$\begin{matrix} 3y+z\equiv 0 & \left( \bmod 23 \right). \\ \end{matrix}$

(2.6)

One can easily to see neither (2.5) nor (2.6) is satisfied for any element $(y, z)\in \{(1, p^{u}q^{v}r^{t}), (p, q^{v}r^{t}), (q, p^{u}r^{t}), (r, p^{u}q^{v}), (pq, r^{t}), (pr, q^{v}), (qr, p^{t}), (pqr, 1), u, v, t=0, 1\}$. This completes the proof of Lemma 2.3.

Lemma 2.4. If $3|b$, then equation (1.7) has no positive integer solution (a, d, y, z) with $ayz-661=b$.

proof. Assume (1.7) has a positive integer solution $(a, d, y, z)$ with $ayz-661=b$. Then we get $b|3y+z$. It implies that $3|z$ since $3|b$, which contradicts with $\gcd(3, z)=1$. This completes the proof of Lemma 2.4.

Lemma 2.5. Let $661+b=2^{k}p$, where k is an integer more than $2$ and p is an odd prime.

If $1+2^{u}p^{v}\not\equiv0\pmod{b}, p+2^{u}\not\equiv0\pmod{b},$ and $2^{t}+p^{v}\not\equiv0\pmod{b}$, where $v\in\{0, 1\}, 0\leq u\leq k, 1\leq t\leq k$, then equation (1.6) has no positive integer solution (a, d, y, z) with $ayz-661=b$.

proof. Assume (1.6) has a positive integer solution $(a, d, y, z)$ with $ayz-661=b$. It follows

$(y, z)\in \{(1, {{2}^{u}}{{p}^{v}}), (p, {{2}^{u}}), ({{2}^{t}}, {{p}^{v}}), ({{2}^{t}}p, 1), 0\le u\le k, v=0, 1, 1\le t\le k\}$

and

$\begin{matrix} y+z\equiv 0 & \left( \bmod b \right), \\ \end{matrix}$

(2.7)

which contradicts the assumption of Lemma 2.5.

This completes the proof of Lemma 2.5.

Lemma 2.6. Let $(a, d, y, z)$ be a positive integer solution of equation (1.6) with $ayz-661=b$. If $b\geq 80, ad\geq 2$, then $y＜6$.

proof. We first prove that $b\leq 1322$. Otherwise $b>1322$, then we have $z＞\frac{y+z}{2}\geq \frac{1322}{2}=661$.

If $ay＞2$, then from $ayz-661=\frac{y+z}{d}\leq 2z$, we get $z\leq (ay-2)z\leq 661$ which contradicts with $z＞661$.

If $a=y=d=1$, then we have $1=-661$ which is impossible.

If $a=y=1, d>1$, then we have

$z=\frac{661d+1}{d-1}=661+\frac{662}{d-1}\le 661+662=1323,$

which contradicts with $z＞1322+661=1983$.

If $a=1, y=2$, then we have

$z=\frac{661}{2}+\frac{665}{2(2d-1)}\le \frac{661+665}{2}=663,$

which contradicts with $z＞1322+661=1983$.

If $a=2, y=1$, then we have

$z=\frac{661}{2}+\frac{663}{2(2d-1)}\le \frac{661+663}{2}=662,$

which also contradicts with $z＞1322+661=1983$. Hence $b\leq 1322$ as desired.

Assume now $y\geq 6$.

If $d\geq 2, b\geq 200$, then $y+z=db\geq 400$. Since quadratic function

$f(y)=y(db-y), 0＜y\le \frac{db}{2},$

is increasing function, so we have

$yz=f(y)\ge f(6)=6\cdot (db-6)\ge 6\cdot 394=2364,$

which contradicts with $ayz\leq 1983$. If $d\geq 2, 80\leq b＜200$, then similarly we have

$yz=f(y)\ge f(6)=6\cdot (db-6)\ge 6\cdot 154=924,$

which contradicts with $ayz\leq 661+200=861$.

If $d=1$, then we have $a\geq 2$. It follows that $yz\leq [\frac{1983}{a}]\leq [\frac{1983}{2}]=991$. If $b\geq 200$, then similarly we have

$yz=f(y)\ge f(6)=6\cdot (b-6)\ge 6\cdot 194=1164,$

which contradicts with $yz\leq 991$. If $80\leq b＜200$, then similarly we have

$yz=f(y)\ge f(6)=6\cdot (b-6)\ge 6\cdot 74=444,$

which contradicts with $yz\leq [\frac{861}{a}]\leq [\frac{861}{2}]=430$.

This completes the proof of Lemma 2.6.

Lemma 2.7. Let (a, d, y, z) be a positive integer solution of equation (1.7) with $ayz-661=b$. If $80\leq b\leq 2644,$ and $a\geq 5$ or $d\geq 7$, then we have either $y＜6$ or $z＜6$.

proof. Since $\gcd(3y, z)=1$, so we have either $3y＜z$ or $3y>z$. We prove that $y＜6$ if $3y＜z$ and $z＜6$ if $3y>z$.

We prove that $z＜6$ if $3y＞z$. Otherwise $z\leq 6$.

Case 1: $d\geq 7$. Then $3y+z=db$, we get $z\leq (ay-2)z\leq 661$ which contradicts with $z＞661$.

If $b\geq 300$, then $y+z=db\geq 2100$. Since quadratic function

$g(z)=z(db-z), 0\text{}z\le \frac{db}{2},$

is increasing function, so we have

$yz=\frac{g(z)}{3}\ge \frac{g(6)}{3}\ge 2\cdot 2094=4188,$

which contradicts with $ayz\leq 3305$.

If $90\leq b＜300$, then similarly we have

$yz=\frac{g(z)}{3}\ge \frac{g(6)}{3}\ge 2\cdot 624=1248,$

which contradicts with $ayz\leq 661+300=961$.

Case 2: $a\geq 5$, then we have $yz\leq [\frac{3305}{a}]\leq [\frac{3305}{5}]=661$.

If $b\geq 400$, then similarly we have

$yz=\frac{g(z)}{3}\ge \frac{g(6)}{3}\ge 2\cdot 394=788,$

which contradicts with $yz\leq 661$.

If $150\leq b＜400$, then similarly we have

$yz=\frac{g(z)}{3}\ge \frac{g(6)}{3}\ge 2\cdot 144=288,$

which contradicts with $yz\leq [\frac{1061}{a}]\leq [\frac{1061}{5}]=212$.

If $90\leq b＜150$, then similarly we have

$yz=\frac{g(z)}{3}\ge \frac{g(6)}{3}\ge 2\cdot 84=168,$

which contradicts with $yz\leq [\frac{811}{a}]\leq [\frac{811}{5}]=162$.

Similarly we can prove that $y＜6$ if $3y＜z$.

This completes the proof of Lemma 2.7.

---

3. Proof of Theorem 1.1

Assume that $(a, d, y, z)$ is a positive integer solution of equation (1.6) with $ayz-661=b\geq 80.$ Then we have $y\leq 5$ by Lemma :2.6. We divide the proof into four cases.

Case 1: $d=1$. From equation (1.6), we get

$\left( ay-1 \right)z=661+y.$

(3.1)

Replacing $y$ by $1, 2, 3, 4, 5$ in (3.1) respectively, we obtain

$(a-1)z=2\cdot 331, (2a-1)z=3\cdot 13\cdot 17, (3a-1)z=8\cdot 83, (4a-1)z=5\cdot 7\cdot 19, (5a-1)z=2\cdot 9\cdot 37.$

Thus we get

$(a, y, z)=(2, 1, 662), (3, 1, 331), (2, 2, 221), (3, 3, 83), (2, 4, 95).$

Case 2: $d=2$. From equation (1.6), we get

$\left( 2ay-1 \right)z=1322+y.$

(3.2)

Replacing $y$ by $1, 2, 3, 4, 5$ in (3.2) respectively, we obtain

$(2a-1)z=9\cdot 49, (4a-1)z=4\cdot 331, (6a-1)z=25\cdot 53, (8a-1)z=6\cdot 13\cdot 17, (10a-1)z=1327.$

Thus we get

$(a, y, z)=(1, 1, 1323), (2, 1, 441), (4, 1, 189), (1, 3, 265).a$

Case 3: $d=3$. From equation (1.6), we get

$\left( 3ay-1 \right)z=1983+y.$

(3.3)

Replacing $y$ by $1, 2, 3, 4, 5$ in (3.3) respectively, we obtain

$(3a-1)z={{4}^{3}}\cdot 31, (6a-1)z=5\cdot 397, (9a-1)z=6\cdot 331, (12a-1)z=1987, (15a-1)z=4\cdot 7\cdot 71.$

Thus we get

$(a, y, z)=(1, 1, 992), (3, 1, 248), (1, 2, 397).$

Case 4: $d\geq 4$. If $b\geq 100$, then by the proof of Lemma 2.6 we know

$yz=f(y)\ge f(6)\ge 6\cdot 394=2364,$

which contradicts with $ayz\leq 1983$. If $80\leq b＜100$, then similarly we know

$yz=f(y)\ge f(6)\ge 6\cdot 314=1884,$

which contradicts with $ayz\leq 761$.

So Theorem 1.1 is proved.

---

4. Proof of Theorem 1.2

Assume that $(a, d, y, z)$ is a positive integer solution of equation (1.7) with $ayz-661=b\geq 90.$ We divide the proof into three cases.

Case 1: $b＞2644$, assume that $y＞z$. Then $y>\frac{3y+z}{4}=\frac{b}{4}>661$. If $az＞4$, then from (1.7), we get $ayz-661=\frac{3y+z}{d}＜4y$. It implies that $y\leq (az-4)y＜661$, which contradicts with $y＞661$. Hence $az\leq 4$. On the other hand, we have $(adz-3)b=az^2+1983\leq 16+1983=1999$ which contradicts with $b>2644$. Similarly we can prove that $y\leq z$ is impossible.

Case 2: $90\leq b\leq 2644, d＜7$ and $a＜5$, then we have $a\in\{1, 2, 3, 4\}, d\in\{1, 2, 4, 5\}$ since $d|3y+z, \gcd(3y, z)=1$.

If $a=1$, from equation (1.7), we get

$(dy-1)(dz-3)=661{{d}^{2}}+3.$

(4.1)

Replacing $d$ by $1, 2, 4, 5$ in (4.1) respectively, we obtain

$(y-1)(z-3)=8\cdot 83, (2y-1)(2z-3)=2647, (4y-1)(4z-3)=71\cdot 149, (5y-1)(5z-3)=16\cdot 1033.$

Thus we get

$(d, y, z)=(1, 2, 667), (1, 3, 335), (1, 5, 169), (1, 9, 86), (1, 84, 11), (1, 167, 7), (1, 333, 5),$

$(1, 665, 4), (2, 1, 1325), (2, 1324, 2), (4, 2645, 1), (5, 1, 827), (5, 1653, 1).$

If $a=2$, from equation (1.7), we get

$(2dy-1)(2dz-3)=1322{{d}^{2}}+3.$

(4.2)

Replacing $d$ by $1, 2, 4, 5$ in (4.2) respectively, we obtain

$(2y-1)(2z-3)=25\cdot 53, (4y-1)(4z-3)=11\cdot 13\cdot 17, (8y-1)(8z-3)=5\cdot 4231, (5y-1)(5z-3)=33053.$

Thus we get

$\begin{align} & (d, y, z)=(1, 1, 664), (1, 3, 134), (1, 27, 14), (1, 133, 4), (1, 663, 2), \\ & (2, 1323, 1), (4, 529, 1). \\ \end{align}$

If $a=3$, from equation (1.7), we get

$(3dy-1)(dz-1)=661{{d}^{2}}+1.$

(4.3)

Replacing $d$ by $1, 2, 4, 5$ in (4.3) respectively, we obtain

$(3y-1)(z-1)=2\cdot 331, (6y-1)(2z-1)=5\cdot {{23}^{2}}, (9y-1)(3z-1)=2\cdot {{5}^{2}}\cdot 7\cdot 17, (15y-1)(5z-1)=2\cdot 8263.$

By computing we get that

$(d, y, z)=(1, 1, 332), (1, 221, 2), (2, 1, 265), (2, 441, 1).$

If $a=4$, we contend $d＜3$. Otherwise $d\geq 4$, we have $3y+z=db\geq 4b$. If $b\geq 300$, then we have $yz\geq 1197$, which contradicts with $yz\leq [\frac{3305}{4}]=826$. If $90\leq b＜300$, then we have $yz\geq 357$, which contradicts with $yz\leq [\frac{961}{4}]=240$. Hence $d＜3$ as desired. From equation (1.7), we get

$(4dy-1)(4dz-3)=2644{{d}^{2}}+3.$

(4.4)

Replacing $d$ by $1, 2$ in (4.4) respectively, we obtain

$(4y-1)(4z-3)=2647, (8y-1)(8z-3)=71\cdot 149.$

Thus we get

$(d, y, z)=(1, 662, 1).$

Case 3: $90\leq b\leq 2644, d\geq 7$ or $a\geq 5$. Then we have either $y\leq 5$ or $z\leq 5$ by Lemma 2.7.

Subcase 1: $d=1, a\geq 5$, from equation (1.7), we get

$\left( ay-1 \right)z=661+3y$

(4.5)

and

$\left( az-3 \right)y=661+z.$

(4.6)

Replacing $y$ by $1, 2, 3, 4, 5$ in (4.5) respectively, we obtain

$(a-1)z=8\cdot 83, (2a-1)z=23\cdot 29, (3a-1)z=2\cdot 5\cdot 67, (4a-1)z=673, (5a-1)z=4\cdot {{13}^{2}}.$

Thus we get

$(a, y, z)=(5, 1, 166).$

Replacing $z$ by $1, 2, 4, 5$ in (4.6) respectively, we obtain

$(a-3)y=2\cdot 331, (2a-3)y=3\cdot 13\cdot 17, (4a-3)y=5\cdot 7\cdot 19, (5a-3)y=2\cdot {{3}^{2}}\cdot 37.$

By computing we get

$(a, y, z)=(5, 331, 1), (8, 51, 2), (10, 39, 2).$

Subcase 2: $d=2, a\geq 5$, from equation (1.7), we get

$\left( 2ay-1 \right)z=1322+y$

(4.7)

and

$\left( 2az-3 \right)y=1322+z.$

(4.8)

Replacing $y$ by $1, 2, 3, 4, 5$ in (4.7) respectively, we obtain

$(2a-1)z={{5}^{2}}\cdot 53, (4a-1)z={{2}^{4}}\cdot 83, (6a-1)z={{11}^{3}}, (8a-1)z=2\cdot 23\cdot 29, (10a-1)z=7\cdot 191.$

By computing we know that the above equations have no positive integer solution $(a, z)$ such that $a\geq 5, b\geq 90$.

Replacing $z$ by $1, 2, 4, 5$ in (4.8) respectively, we obtain

$(2a-3)y={{3}^{3}}\cdot {{7}^{2}}, (4a-3)y={{2}^{2}}\cdot 331, (8a-3)y=2\cdot 3\cdot 13\cdot 17, (10a-3)y=1327.$

By computing we get

$(a, y, z)=(6, 147, 1), (12, 63, 1), (5, 189, 1).$

Subcase 3: $d\geq 4, a\geq 5$, then we have $3y+z=db\geq 4b$. If $b\geq 200$, then we get that $yz\geq 797$ which contradicts with $yz\leq [\frac{3305}{5}]=661$. If $90\leq b＜200$, then we get that $yz\geq 357$ which contradicts with $yz\leq [\frac{861}{5}]=172$.

Subcase 4: $d\geq 7, a＜5$. We contend that $a=1$. Otherwise $a\geq 2$. If $b\geq 300$, then we get that $yz\geq 2097$ which contradicts with $yz\leq [\frac{3305}{2}]=1652$. If $90\leq b＜300$, then we get that $yz\geq 627$ which contradicts with $yz\leq [\frac{861}{2}]=430$. Thus $a=1$ as desired. From equation (1.7), we get

$(dy-1)(yz-661)=661+3{{y}^{2}}$

(4.9)

and

$(dz-3)(yz-661)=1983+{{z}^{2}}.$

(4.10)

Replacing $y$ by 1, 2, 3, 4, 5 in (4.9) respectively, we obtain

$\begin{align} & (d-1)(z-661)=8\cdot 83, (2d-1)(2z-661)=673, (3d-1)(3z-661)={{4}^{2}}\cdot 43, (4d-1)(4z-661)=709, \\ & (5d-1)(5z-661)={{2}^{5}}\cdot 23. \\ \end{align}$

By computing we know that the above equations have no positive integer solution $(a, z)$ such that $d\geq 7, b\geq 90$.

Replacing $z$ by $1, 2, 4, 5$ in (4.8) respectively, we obtain

$(d-3)(y-661)={{2}^{6}}\cdot 31, (2d-3)(2y-661)=1987, (4d-3)(4y-661)=1999, (5d-3)(5y-661)=8\cdot 251.$

By computing we get that

$(d, y, z)=(7, 1157, 1), (11, 909, 1), (19, 785, 1).$

Therefore the proof of Theorem 1.2 is complete.

---

5. Proof of Theorem 1.3

Since

$\begin{align} & \begin{matrix} 661+12=673\equiv 1 & \left( \bmod 3 \right), 661+22=683\equiv 1 & \left( \bmod 11 \right), 661+30=691\equiv 1 & \left( \bmod 3 \right), \\ \end{matrix} \\ & \begin{matrix} 661+40=701\equiv 1 & \left( \bmod 5 \right), 661+48=709\equiv 1 & \left( \bmod 3 \right), 661+58=719\equiv 23 & \left( \bmod 29 \right), \\ \end{matrix} \\ & \begin{matrix} 661+66=727\equiv 1 & \left( \bmod 3 \right), 661+72=733\equiv 1 & \left( \bmod 3 \right), 661+78=739\equiv 1 & \left( \bmod 3 \right), \\ \end{matrix} \\ & \begin{matrix} 661+82=743\equiv 5 & \left( \bmod 41 \right), \\ \end{matrix} \\ \end{align}$

so both (1.6) and (1.7) have no positive integer solution $(a, d, y, z)$ with $ayz-661=b$ for $b\in\{12, 22, 30, 40, 48, 58, 66, 72, 78, 82\}$ by Lemma 2.1.

Since

$\begin{align} & \begin{matrix} 661+10=671=11\cdot 61, 11\equiv 61\equiv 1 & \left( \bmod 5 \right), \\ \end{matrix} \\ & \begin{matrix} 661+20=681=3\cdot 227, 227\equiv 7 & \left( \bmod 20 \right), 661+26=687=3\cdot 229, 229\equiv -5 & \left( \bmod 13 \right), \\ \end{matrix} \\ & 661+28=689=13\cdot 53, \\ & \begin{matrix} 661+34=695=5\cdot 139, 139\equiv 3 & \left( \bmod 17 \right), \\ \end{matrix} \\ & \begin{matrix} 661+37=698=2\cdot 347, 347\equiv 14 & \left( \bmod 37 \right), \\ \end{matrix} \\ & \begin{matrix} 661+38=699=3\cdot 233, 233\equiv 5 & \left( \bmod 19 \right), \\ \end{matrix} \\ & \begin{matrix} 661+46=707=7\cdot 101, 101\equiv 9 & \left( \bmod 23 \right), \\ \end{matrix} \\ & 661+52=713=23\cdot 31, \\ & \begin{matrix} 661+56=717=3\cdot 239, 239\equiv 1 & \left( \bmod 7 \right), \\ \end{matrix} \\ & \begin{matrix} 661+57=718=2\cdot 359, 359\equiv 17 & \left( \bmod 57 \right), \\ \end{matrix} \\ & \begin{matrix} 661+60=721=7\cdot 103, 7\equiv 103\equiv 1 & \left( \bmod 3 \right), \\ \end{matrix} \\ & \begin{matrix} 661+62=723=3\cdot 241, 241\equiv -7 & \left( \bmod 31 \right), \\ \end{matrix} \\ & \begin{matrix} 661+70=731=17\cdot 43, 17\equiv 3 & \left( \bmod 7 \right), 43\equiv 1 & \left( \bmod 7 \right), \\ \end{matrix} \\ & \begin{matrix} 661+56=717=3\cdot 239, 239\equiv 15 & \left( \bmod 56 \right), \\ \end{matrix} \\ & \begin{matrix} 661+73=734=2\cdot 367, 367\equiv 2 & \left( \bmod 73 \right), \\ \end{matrix} \\ & 661+76=737=11\cdot 67, \\ & \begin{matrix} 661+85=746=2\cdot 373, 373\equiv 33 & \left( \bmod 85 \right), \\ \end{matrix} \\ & \begin{matrix} 661+88=749=7\cdot 107, 107\equiv 19 & \left( \bmod 88 \right), \\ \end{matrix} \\ \end{align}$

so both (1.6) and (1.7) have no positive integer solution $(a, d, y, z)$ with $ayz-661=b$ for $b\in\{10, 20, 26, 28, 34, 37, 38, 46, 52, 56, 57, 60, 62, 70, 73, 76, 85, 88\}$ by Lemma 2.2 (i).

Since

$\begin{align} & \begin{matrix} 661+31=692={{2}^{2}}\cdot 173, 173\equiv 18 & \left( \bmod 31 \right), \\ \end{matrix} \\ & \begin{matrix} 661+44=705=3\cdot 5\cdot 47, 47\equiv 3 & \left( \bmod 11 \right), \\ \end{matrix} \\ & \begin{matrix} 661+50=711={{3}^{2}}\cdot 79, 79\equiv 4 & \left( \bmod 25 \right), \\ \end{matrix} \\ & \begin{matrix} 661+55=716={{2}^{2}}\cdot 179, 179\equiv 14 & \left( \bmod 55 \right), \\ \end{matrix} \\ & 661+61=722=2\cdot {{19}^{2}}, \\ & 661+64=725={{5}^{2}}\cdot 29, \\ & 661+80=741=3\cdot 13\cdot 19, \\ \end{align}$

so both (1.6) and (1.7) have no positive integer solution $(a, d, y, z)$ with $ayz-661=b$ for $b\in\{31, 44, 50, 55, 61, 64\}$ by Lemma 2.3 (i).

Since

$\begin{align} & 661+47=708={{2}^{2}}\cdot 3\cdot 59, 661+59=720={{2}^{4}}\cdot {{3}^{2}}\cdot 5, 661+68=729={{3}^{6}}, \\ & 661+71=732={{2}^{2}}\cdot 3\cdot 61, 661+77=738=2\cdot {{3}^{2}}\cdot 41, \\ \end{align}$

by computing we know both (1.6) and (1.7) have no positive integer solution $(a, d, y, z)$ with $ayz-661=b$ for $b\in\{47, 59, 68, 71, 77\}$.

Since $661+27=688=2^4\cdot43, 27\nmid x,$ for any

$x\in \{1+{{2}^{\alpha }}{{43}^{\beta }}, 43+{{2}^{\alpha }}, {{2}^{t}}+{{43}^{\beta }}\},$

where $\alpha\in\{0, 1, 2, 3, 4\}, \beta\in\{0, 1, \}, t\in\{1, 2, 3, 4\}.661+75=736=2^5\cdot23, 75\nmid x,$ for any

$x\in \{1+{{2}^{\alpha }}{{23}^{\beta }}, 23+{{2}^{\alpha }}, {{2}^{t}}+{{23}^{\beta }}\},$

where $\alpha\in\{0, 1, 2, 3, 4, 5\}, \beta\in\{0, 1, \}, t\in\{1, 2, 3, 4, 5\}$ so (1.6) has no positive integer solution $(a, d, y, z)$ with $ayz-661=b$ for $b\in\{27, 75\}$ by Lemma 2.5.

Since

$\begin{align} & \begin{matrix} 661+18=679=7\cdot 97, 7\equiv 97\equiv -2 & \left( \bmod 9 \right), \\ \end{matrix} \\ & \begin{matrix} 661+24=685=5\cdot 137, 5\equiv 137\equiv 1 & \left( \bmod 4 \right), \\ \end{matrix} \\ & \begin{matrix} 661+33=694=2\cdot 347, 347\equiv 6 & \left( \bmod 11 \right), \\ \end{matrix} \\ & \begin{matrix} 661+36=697=17\cdot 41, 17\equiv 41\equiv 1 & \left( \bmod 4 \right), \\ \end{matrix} \\ & \begin{matrix} 661+42=703=19\cdot 37, 19\equiv 37\equiv 1 & \left( \bmod 3 \right), \\ \end{matrix} \\ & \begin{matrix} 661+45=706=2\cdot 353, 353\equiv 2 & \left( \bmod 9 \right), \\ \end{matrix} \\ \end{align}$

so (1.6) has no positive integer solution $(a, d, y, z)$ with $ayz-661=b$ for $b\in\{18, 24, 33, 36, 42, 45\}$ by Lemma 2.2 (ii).

For any $b=3t, 1\leq t＜29$, (1.7) has no positive integer solution $(a, d, y, z)$ with $ayz-661=b$ by Lemma 2.4.

Since

$\begin{align} & 661+3={{2}^{3}}\cdot 83, 661+6=2\cdot {{3}^{2}}\cdot 37, 661+9=2\cdot 5\cdot 67, 661+15={{2}^{2}}\cdot {{13}^{2}}, \\ & \\ \end{align}$

$661+21=2\cdot 11\cdot 31, 661+39={{2}^{2}}\cdot {{5}^{2}}\cdot 7, 661+51={{2}^{3}}\cdot 89,$

so by computing we get that $(a, d, y, z)=$

(2, 29, 4, 83), (2, 112, 1, 332), (8, 28, 1, 83), (83, 3, 1, 8), (332, 1, 1, 2), (23, 5, 1, 29),

(29, 4, 1, 23), (2, 8, 5, 67), (5, 15, 1, 134), (26, 1, 2, 13), (2, 2, 11, 31), (11, 3, 1, 62),

(2, 1, 14, 25), (5, 1, 4, 35), (2, 7, 1, 356),

are positive integer solutions of (1.6).

We are now in a position to discuss

$b\in \{1, 2, 4, 5, 7, 8, 11, 13, 14, 16, 17, 19, 23, 25, 28, 29, 32, 35, 41, 43, 49, 53, 59, 65, 67, 74, 79\}.$

Since

$\begin{align} & 661+1=2\cdot 331, 661+2=3\cdot 13\cdot 17, 661+4=5\cdot 7\cdot 19, 661+5=2\cdot {{3}^{2}}\cdot 37, \\ & 661+7={{2}^{2}}\cdot 167, 661+8=3\cdot 223, 661+11={{2}^{5}}\cdot 3\cdot 7, 661+13=2\cdot 337, 661+14={{3}^{3}}\cdot {{5}^{2}} \\ & 661+16=677, 661+17=2\cdot 3\cdot 113, 661+19={{2}^{3}}\cdot 5\cdot 17, 661+23={{2}^{2}}\cdot {{3}^{2}}\cdot 19, \\ & 661+25=2\cdot {{7}^{3}}, 661+28=13\cdot 53, 661+29=2\cdot 3\cdot 5\cdot 23, 661+32={{3}^{2}}\cdot 7\cdot 11, \\ & 661+35={{2}^{3}}\cdot 3\cdot 29, 661+41=2\cdot {{3}^{3}}\cdot 13, 661+43={{2}^{6}}\cdot 11, 661+49=2\cdot 5\cdot 71, \\ & 661+53=2\cdot 3\cdot 7\cdot 17, 661+59={{2}^{4}}\cdot 5\cdot {{3}^{2}}, 661+65=2\cdot 3\cdot {{11}^{2}}, \\ & 661+67={{2}^{3}}\cdot 7\cdot 13, 661+74=3\cdot 5\cdot {{7}^{2}}, 661+79={{2}^{2}}\cdot 5\cdot 37, \\ \end{align}$

so by computing we get that $(a, d, y, z)=$

(662, 2, 1, 1), (1, 663, 1, 662), (1, 333, 2, 331), (2, 332, 1, 331), (331, 3, 1, 2), (663, 1, 1, 1),

(1, 332, 1, 663), (1, 112, 3, 221), (1, 32, 13, 51), (1, 28, 17, 39), (3, 111, 1, 221), (3, 14, 13, 17),

(13, 26, 1, 51), (13, 10, 3, 17), (17, 20, 1, 39), (17, 8, 3, 13), (39, 9, 1, 17), (51, 7, 1, 13),

(221, 2, 1, 3), (7, 6, 5, 19), (19, 3, 5, 7), (7, 24, 1, 95), (19, 9, 1, 35), (35, 5, 1, 19),

(95, 2, 1, 7), (1, 11, 18, 37), (1, 67, 2, 333), (9, 15, 1, 74), (74, 2, 1, 9), (6, 8, 3, 37),

(111, 1, 2, 3), (4, 24, 1, 167), (3, 28, 1, 223), (32, 2, 1, 21), (24, 1, 4, 7), (28, 1, 3, 8),

(21, 3, 1, 32), (2, 26, 1, 337), (15, 1, 5, 9), (9, 2, 3, 25), (25, 2, 1, 27), (1, 7, 6, 113),

(2, 20, 1, 339), (1, 3, 17, 40), (20, 1, 2, 17), (9, 1, 4, 19), (14, 2, 1, 49), (5, 1, 6, 23),

(6, 4, 1, 115), (11, 2, 1, 63), (3, 1, 11, 21), (4, 1, 6, 29), (4, 5, 1, 174), (9, 1, 2, 39),

(2, 1, 11, 32), (1, 3, 5, 142), (7, 1, 2, 51), (5, 2, 1, 147), (2, 1, 5, 74)

are the solutions of (1.6) and $(a, d, y, z)=$

(662, 4, 1, 1), (1, 337, 2, 331), (1, 665, 1, 662), (1, 1987, 662, 1), (1, 995, 331, 2), (2, 334, 1, 331),

(2, 994, 331, 1), (331, 5, 1, 2), (331, 7, 2, 1), (663, 2, 1, 1), (1, 333, 1, 663), (1, 995, 663, 1),

(1, 115, 3, 221), (1, 83, 51, 13), (1, 67, 39, 17), (3, 112, 1, 221), (3, 332, 221, 1), (3, 28, 13, 17),

(3, 32, 17, 13), (13, 13, 3, 17), (17, 59, 39, 1), (17, 11, 3, 13), (39, 10, 1, 17), (39, 26, 17, 1),

(51, 8, 1, 13), (51, 20, 13, 1), (221, 5, 3, 1), (1, 167, 1, 665), (665, 1, 1, 1), (1, 37, 5, 133),

(1, 101, 133, 5), (1, 31, 35, 19), (1, 73, 95, 7), (5, 34, 1, 133), (5, 10, 7, 19), (5, 16, 19, 7),

(133, 2, 1, 5), (5, 100, 133, 1), (333, 1, 1, 2), (18, 8, 1, 37), (222, 2, 3, 1), (3, 11, 6, 37),

(3, 67, 111, 2), (37, 11, 18, 5), (2, 200, 333, 1), (334, 1, 2, 1), (167, 1, 1, 4), (1, 29, 3, 223),

(6, 5, 16, 7), (8, 23, 84, 1), (84, 1, 1, 8), (96, 2, 7, 1), (112, 1, 3, 2), (45, 1, 3, 5),

(27, 2, 1, 25), (75, 2, 9, 1), (1, 127, 677, 1), (3, 7, 2, 113), (6, 20, 113, 1), (34, 1, 5, 4),

(4, 2, 9, 19), (13, 2, 1, 53), (15, 1, 2, 23), (9, 1, 7, 11), (33, 2, 21, 1), (12, 1, 2, 29),

(12, 5, 58, 1), (3, 2, 4, 58), (27, 1, 13, 2), (26, 2, 27, 1), (8, 1, 18, 5), (2, 2, 3, 121),

(2, 1, 13, 28), (15, 2, 49, 1)

are the solutions of (1.7).

We are now in a position to discuss $b\in\{83, 86, 89\}.$

Since $661+83=2^3\cdot3\cdot31$, $661+86=3^2\cdot83$ so by computing we get that $(a, d, y, z)=(1, 9, 1, 744), (9, 1, 1, 83)$ are solutions of (1.7).

Since $661+89=2\cdot3\cdot5^3$, so by computing we get that (1.7) has no positive integer solution.

This finishes the proof of Theorem 1.3.

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6. Proof of Theorem 1.4

If $ay＞2$, then from $ayz-661=\frac{661y+z}{d}＜2z$, we get that $z\leq (ay-2)z＜661$ which contradicts with $z＞661y$. Hence $a=y=1$ or $a=1, y=2$ or $a=2, y=1$.

$a=y=1$ implies that $(d-1)(z-661)=2\cdot661$. So

$(d, z)=(2, 1983), (3, 1322), (662, 663), (1323, 662).$

$a=1, y=2$ implies that $(2d-1)(2z-661)=5\cdot661$. So

$(d, z)=(1, 1983), (3, 661), (331, 333), (1653, 331).$

$a=2, y=1$ implies that $(2d-1)(2z-661)=3\cdot661$. So

$(d, z)=(1, 1322), (2, 661), (331, 332), (992, 331).$

This concludes the proof of Theorem 1.4.

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Conflict of Interest

All authors declare no conflicts of interest in this paper.

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