<s>[INST] Given $a,b$, find $c$ such that $a+c, b+c, ab+c$ are perfect squares 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.
 [/INST] 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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