perfect squares possible? If we let a, b, c, d, and x be integers is it possible that $$x^2+a^2 = (x+1)^2 + b^2 = (x+2)^2 + c^2 = (x+3)^2 + d^2$$
My initial thought is no way! I tried expanding and simplifying, getting $$a^2 = 2x+1 + b^2 = 4x+4 + c^2 = 6x+9 + d^2$$.
It seems that the difference between these perfect squares is impossible - but why? Consecutive perfect squares always differ by some $2n+1$ term, and each term must continue to increase by an odd amount...
Any ideas in the right direction are greatly appreciated!

I get it. Mod 4 suffices. Well, mod 8 anyway. One of $x,x+1,x+2,x+3$ is divisible by 4. If its matching partner out of $a,b,c,d$ is even, the common sum is divisible by 4, and we actually cannot have any odd numbers involved, which cannot occur with consecutive $x,x+1,x+2,x+3.$ Required detail: the sum of two odd squares is $2 \pmod 4.$
So, the partner is odd, the sum is $1 \pmod 8.$ One of the other $x+j$ numbers is $2 \pmod 4,$ so that sum must be $5 \pmod 8.$ So, actually, impossible.
Done.
