# 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!

• Perhaps reducing this mod 4 or something similar may be one way to approach the problem. Oct 29, 2013 at 20:42
• You can get three in a row, $$1105 = 31^2 + 12^2 = 32^2 + 9^2 = 33^2 + 4^2.$$ My first impression is that four ought to be possible. Try $n = 5 \cdot 13 \cdot 17 \cdot 29 = 32045,$ then try $n = 5 \cdot 13 \cdot 17 \cdot 29 \cdot 37 = 1185665.$ And so on, multiply by the next prime $4k+1.$ Meanwhile, where did you get this problem?? Oct 29, 2013 at 20:48
• @WillJagy I have tried that approach in pari/gp, and no combination seems to work, unfortunately :( Oct 29, 2013 at 20:51
• @OldJohn, sometimes pari gets these problems that they call kidney stones, it cannot output what you want it to. Oct 29, 2013 at 20:53
• The problem comes from an old sheet of basic number theory practice problems - all intended to be able to solve by hand. I thought perhaps I was missing something obvious... Oct 29, 2013 at 21:00

## 1 Answer

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.