I got this question online:
Find infinitely many triplets of naturals $a,b,c$ in arithmetibc progression such that $ab+1,bc+1,ac+1$ are perfect squares.
My attempt:
Let $a=b-d,c=b+d$
So
$b^2-bd+1=j^2$
$b^2+bd+1=k^2$
$b^2-d^2+1=l^2$
This gives
$k^2-j^2=2bd$
This implies that $j,k$ cannot be consecutive squares, and also that $bd$ is even.
Also, I get $d=\sqrt{j^2+k^2-2l^2\over2}$. And I tried substituting $j^2+k^2=2b^2+2$, but it doesn't seem this will go anywhere.
Please help.
This is similar, but doesn't have the AP condition.