Find all positive integers $d$ such that $d$ divides both $n^{2}+1$ and $(n + 1)^{2}+1$ for some integer $n$.

Currently what I am thinking of is like manipulating $n^{2}+1$ and finding out the answer. Please guide me on how to move forward with this problem.


Thanks for helping.

  • 2
    $\begingroup$ Hint: $d$ divides their difference as well ! $\endgroup$
    – r9m
    Dec 27, 2014 at 10:14
  • $\begingroup$ Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ Dec 27, 2014 at 13:08

1 Answer 1


The idea is to reach at a constant divisible by $d$

If $d$ divides $a,b;d$ must divide $ax+by$ for integers $a,b,x,y$

So, $d$ must divide $1\cdot(n+1)^2+1(-1)\cdot(n^2+1)=2n+1$

So, $d$ must divide $n\cdot(2n+1)+(-2)\cdot(n^2+1)=n-2$

Follow the pattern to find $d$ must divide $5$

  • $\begingroup$ That is more than the hint asked for $\endgroup$ Dec 27, 2014 at 10:17
  • $\begingroup$ @DevarshRuparelia, Was it more than hint? $\endgroup$ Dec 27, 2014 at 10:20
  • 1
    $\begingroup$ (+1) .. also I think the downvote is a bit harsh in this context ! $\endgroup$
    – r9m
    Dec 27, 2014 at 10:21
  • $\begingroup$ You basically showed how to solve the problem, that is more than a hint. $\endgroup$ Dec 27, 2014 at 10:22
  • $\begingroup$ @labbhattacharjee I think so but not too much that it can be downvoted.I would say remove the last step and it would be fine. $\endgroup$ Dec 27, 2014 at 10:22

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