Find all integers $m,n$ such that $(562-5(m+n))(82-5(m-n))=98^2$

Question

Find all integers $m,n$ such that $$(562-5(m+n))(82-5(m-n))=98^2.$$

What is a simpler method to tackle this? I know the results but my approach is kind of lengthy. I considered every possible factor of $98^2$ to get the result. Thanks.

Answer 1

$$(562-5S)(82-5D)=98^2=2^2\cdot7^4$$ The number $98^2$ has $15$ divisors, $$d\in\{1,2,4,7,14,28,49,98,196,343,686,1372,2401,4802,9604\}$$ Because of $82-5D=d\iff82-d=5D$, it follows that the only divisors to be considered are $d\in\{2,7,1372,4802\}$.

It follows the possibilities $$(D,S)\in\{(16,-848),(15,-162),(-162,15),(-848,16)\}$$

From which the only solution $\color{red}{(m,n)=(-416,-432)}$

Answer 2

From $562-5(m+n)=a, 82-5(m-n)=b$ one gets $m=\frac {644 -a -b}{10}, n=\frac {480-a + b}{10}$ therefore $10 | (a-b)$ where $a\cdot b = 98^2$. There are 30 factorizations of $98^2$ over integers, just filter out those not satisfying $10 | (a-b)$ condition, you'll only get a few remaining.

Answer 3

Let $a$ be $562 - 5(m + n)$ $=$ $a$. Let $b$ be $82 - 5(m - n)$ $=$ $b$.

Now, you check if the last digit of $a$ has $2$ or $-8$ because it has to be $(a - 562)$ % $5$ $=$ $0$ where $m, n$ $\in$ $Z$. Therefore you should check just $2$ or $-28$, $-98$, $1372$, $4802$ are one of the factor from the combinations of $(2^2)$$・$$(7^4)$

And, where $a$: $-28$ : ($-2^2$・$7$), $b$: $-7^3$, where $a$: $1372$:($2^2$・$7^3$) $b$: $7$.

Now, $(-7^3 - 82)$ % $5$ $\ne$ $0$ and whrere $a = 1372$ and $b = 7$. = $0$ will not make $m$, $n$ $\notin$ $Z$ since $m = \frac{177}{2}$, $n = \frac{147}{2}$ . So, you only need to consider the $a = 2, -98, 4802.$ The answers are $(m, n) = (84, 48)$$[a = -98, b = -98]$, $(-416, 528)$$[a = 2, b = 4802]$, $(-416, -432)$$[a = 4802, b = 2]$.