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I have a function $f(x) = 4x^2 + 4ax + b$ where $a$ and $b$ are both even numbers. I also have an interval within which there are only two integers $x$ that will result in a perfect square. $a$, $b$, and the larger of the two perfect squares are known.

Is there some way that does not involve prime factorization to find or get a rough estimate of what value $x$ will result in the smaller perfect square?

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Let us call these two values of $x, \ x_1$ and $x_2$, being the larger of the two. Then as $x_2$ is known, we can write that $x_1=x_2-m$ for some integer $m$. Then we get that we know $f(x_1)$ to be a perfect square we get that $f(x_2-m)$ is also a perfect square. $$f(x_2-m)=4(x_2-m)^2+4a(x_2-m)+b=4m^2-(8x_2+4a)m+(4x_2^2+4ax_2+b)$$ which is a quadratic in $m$. Solve this for $m$ to be in your known interval and you have $x_1$

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  • $\begingroup$ Maybe I'm missing something, but don't quadratic equations work on the assumption that the equation equals 0? Since this equals the perfect square I'm looking for, I would have to subtract that square from the equation first. That means I'd have to know the answer to find the answer, no? $\endgroup$ – strogre May 9 at 22:30
  • $\begingroup$ If it is a perfect square, you can complete the square as it were, and you know that the remaining term is 0 $\endgroup$ – W M Seath May 9 at 22:43

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