<s>[INST] divisibility problem and gcd Let a and b be a positive integers. Proof that if number $ 100ab -1 $ divides number $ (100a^2 -1)^2 $ then also divides number $ (100b^2 -1)^2 $.
My attempt:
Let's notice that \begin{split} b^2(100a^2-1)^2 -a^2(100b^2-1)^2 & =(100a^2b-b)^2-(100ab^2-a)^2\\
 & =(100a^2b-b-100ab^2+a)(100a^2b-b+100ab^2-a)\\
 & =(100ab(a-b)+(a-b))(100ab(a+b)-(a+b))\\
 & =(a-b)(100ab+1)(a+b)(100ab-1).\end{split}
This means that $ 100ab-1 |a^2(100b^2-1)^2$, so if we proof that $\gcd(100ab-1,a^2)=1$ the proof is completed. Now I know that it  should be trivial to show that these numbers are relatively prime but somehow i have no idea how to do it.
Also I am intrested if there is a way to solve this problem by using modular arithmetic?
 [/INST] For a modular arithmetic argument:
Let $N=100ab-1$ and work modulo $N$.
Starting with $(100a^2 -1)^2\equiv 0$,  multiply by $10^4b^4$:
$$(10^4a^2b^2-100b^2)^2\equiv 0.$$
Since $10^2ab\equiv 1$, we have $(1-100b^2)^2\equiv 0.$
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