Given that $x, y$ are positive integers with $x(x + 1)\mid y(y + 1)$, but neither $x$ nor $x + 1$ divides either of $y$ or $y + 1$, and $x^2+ y^2$ as small as possible, find $x^2+ y^2$.

I have tried looking at the values, and it seems that neither $x$ or $x+1$ or the $y$'s are prime.

  • $\begingroup$ I don't see the meaning of $j$? Please correct your post. $\endgroup$ – BIS HD Oct 29 '13 at 0:54
  • $\begingroup$ Could someone please change $x2+y2$ to a sum of squares? My edit won't be processed. $\endgroup$ – Yadnarav3 Oct 29 '13 at 1:03

Yes, we don't want $x$ or $x+1$ to be a prime power. The first candidate is $x=14$. Then $y=20$ works.

Added: Suppose we find consecutive composites integers $x$ and $x+1$ neither of which is a prime power. Let $x=ab$ and $y=cd$, where $a$ and $b$ are relaively prime, as are $c$ and $d$, and none is equal to $1$. Consider the system of congruences $y\equiv 0\pmod{ac}$, $y=\equiv -1\pmod{bd}$. By the Chinese Remainder Theorem, this has a solution. Note that $x$ does not divide $y$, for $b$ divides $y+1$, so is relatively prime to $y$. The other required "non-divisibilities" can be verified in a similar way. But $x(x+1)$ divides $y(y+1)$.

  • $\begingroup$ How can we reach the conclusion without guessing? $\endgroup$ – Yadnarav3 Oct 29 '13 at 16:14
  • $\begingroup$ Looking at particular numbers is not guessing, it is finding out what's going on. Nothing serious gets done without preliminary exploration of the territory, sometimes taking years. And I knew that from $2$ consecutive integers that are not prime powers, I could construct suitable $y$ by a Chinese Remainder Theorem argument. Then I would have an upper bound, and the rest in principle would be bounded search. That part I would not have done, it is not interesting. But it so happened that there was a small example. One can produce larger ones, but the question asked for a minimum. $\endgroup$ – André Nicolas Oct 29 '13 at 16:26
  • $\begingroup$ Thank you. If you don't mind answering, can you also please show me the Chinese remainder argument for a bound? Otherwise, feel free to disregard this comment. $\endgroup$ – Yadnarav3 Oct 29 '13 at 20:06
  • $\begingroup$ I have added the sketch of a Chinese Remainder Theorem argument. The $y$ mentioned can be taken to be less than $x(x+1)$. $\endgroup$ – André Nicolas Oct 29 '13 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.