# 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.

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

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:$$a+c=a+(k^2-a)=k^2\\b+c=(a+2k+1)+(k^2-a)=k^2+2k+1=(k+1)^2\\ab+c=a(a+2k+1)+(k^2-a)=a^2+2ka+a+k^2-a=a^2+2ka+k^2=(a+k)^2$$