Find all integers $m,n$ such that $$(562-5(m+n))(82-5(m-n))=98^2.$$
Find all integers $m,n$ such that $(562-5(m+n))(82-5(m-n))=98^2$
I have my answer to the question above. Now, my question is there are four different interpretations to this question. I have my answer which satisfies the equation. Although other users' answers are more beautiful and persuading, and they have different answers to $(m, n)$. It's like the movie Rashomon. What should I rely upon?
I solved and found $(m, n) = (84, 48)$, $(-416, 528)$, $(-416, -432)$ considering $30$ diviors of $98^2$ in the way described below. But one user state the answer is $(m, n)$ = $(-416, -432).$ considering $15$divisors of $98^2$ ising his method similar to mine. However, other user states "$562-5(m+n)=a, 82-5(m-n)=b$ $\Rightarrow$ $m=\frac {644 -a -b}{10}, n=\frac {480-a + b}{10}$ 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" In this case, this one is the most upvoted answer though, I don't know why I don't have to consider $(a + b) $%$ 10$ $=$ $4$. Furthermore, the other user state using Mathematica $10.0$(powerful software) gives $m = -416, n = -432.$.
I would like to this question since I have already asked the other users' could provide the reason why we have different answers. Unfortunately, I assume probably they might be too busy to respond it.
Thank you.
Appendix: My answer to the original post.
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]$.
Your definition of the two parts of the LHS of the original equation is useful, $a:= 562-5(m+n)$ and $b:=82-5(m-n) $ with $ab=98^2$.
And your answer is correct. For another view of the process, I have laid out my approach which is slightly different again, reducing the possibilities to consider wherever possible.
You can see that $(m+n)$ and $(m-n)$ (and thus $a$ and $b$) are either both odd or both even, and we need at least one of $a,b$ even so both must be. Since $98=2\cdot 7^2$, this reduces the factor pairs to consider down to just $(98,98)$, $(14,686)$ and $(2,4802)$ and their negative counterparts. Considering the requirement to have both factors $\equiv 2 \bmod 10$, from the form of $a$ and $b$, we reduce further to $(-98,-98)$ and $(2,4802)$ (in either order).
These will define the three solutions as you can calculate $(m,n)$ directly from these:
\begin{array}{c|r} (a,b) & m+n & m-n & m & n \\ \hline (-98,-98) & 132 & 36 & 84 & 48 \\[1pt] (2,4802) & 112 & -944 & -416 & 528 \\[1pt] (4802,2) & -848 & 16 & -416 & -432 \\[1pt] \end{array}