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I am asking if there exists any method to determine all positive integers $a$ and $b$ such that

$(4b+3)^2$ divide $4a+8ab^2+12ab+1$

My solution: For $b=0$, we have $9$ divide $4a+1$. One can choose $a=2$. However, I am not able to go further with this method.

Safwane
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  • If you use the slight generalization of the Euclidean algorithm described in my answer in the linked dupe (considering it a a polynomial in $b),,$ then you end up doing the same computation as in Thomas' answer (i.e. that answer is a special case of the linked method). – Bill Dubuque Jul 01 '21 at 00:31

1 Answers1

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Since $(4b+3)^2$ is odd, it dives a number if and only if it divides $2$ times that number.

But $$2(4a+8ab^2+12ab+1)=a(4b+3)^2-a+2.$$

So you need $$a\equiv 2\pmod {(4b+3)^2}\tag1$$

In particular, $a=2$ always works, but any $a$ satisfying $(1)$ works.

Thomas Andrews
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