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Prove that for every different parity numbers $a,b \in \Bbb N$ there exist $c \in \Bbb Z$ such that numbers $a+c, b+c, ab+c$ are perfect squares.

I tried to find separate solutions, $a=2, b=7, c=2$ also $a=6, b=13, c=3$ and etc., then all $a+c, b+c, ab+c$ are perfect squares. But have no idea how to start proof in general.

Have any ideas how to start proof? Thanks.

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    $\begingroup$ What have you tried? ${}{}{}$ $\endgroup$ Dec 10, 2019 at 19:57
  • $\begingroup$ @ThomasAndrews updated $\endgroup$
    – Spameris
    Dec 10, 2019 at 20:11

1 Answer 1

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Since $b$ is of a different parity than $a$, we may choose an integer $k$ such that $b=a+2k+1$. Then take $c=k^2-a$. Then we have:$$a+c=a+(k^2-a)=k^2\\b+c=(a+2k+1)+(k^2-a)=k^2+2k+1=(k+1)^2\\ab+c=a(a+2k+1)+(k^2-a)=a^2+2ka+a+k^2-a=a^2+2ka+k^2=(a+k)^2$$

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