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Like given some multiple of 16, (in the integers), is it possible to show that there is some multiple of 4 that would make it a perfect square? For example, 32, you can add 4 to make it a perfect square? Can we prove this? If yes, then how, because I'm stuck. Or do we need more info (because there is a second (somewhat separate) condition I'm leaving out because it doesn't seem useful. Any help would be appreciated. I'm so stuck

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If I interpret your question correctly, then given $16k$, does there exist $n \in \Bbb{N}$ such that $4n+16k$ is a perfect square?

Let $n=k^2+4$, then $4n+16k=4(k^2+4)+16k=(2k+4)^2$

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    $\begingroup$ How did you come up with this??? This is genius! Omg. Any tips? Cause I need to do similar stuff, replacing the multiple of 16 with other numbers like 4j +1 etc.) So I kind of want to understand how you came up with this $\endgroup$ – Chubbles Oct 26 '18 at 2:37
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If $k$ is given, then take $n$ sufficiently large such that $n^2-4k >0$. Then $16k + 4(n^2-4k)=(2n)^2$ is a perfect square.

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