Assume $p \in \mathbb P.$ Assume $0<p-2k<p$ and the next square larger than $p(p-2k)$ is $(p-k)^2$.

It is trivial to show that $p(p-2k)+k^2$ is a square. Simply $p(p-2k)+k^2 = (p-k)^2.$ Starting with $k = 1$ we get next square larger than $p(p-2k)$ is $(p-k)^2$ and the distance to it is $k^2$.

However, at some point this fails to be true. For example, take $p = 41$. Then $41(41-2*8) + 8^2 = 33^2$ and similarly for all $k$ smaller than 8. But $41(41-2*9) = 943$ and $943 + 9^2 = 32^2 > 31^2 = 943 + 18$. Thus the next square larger than $41(41-2*9)$ is not $(41-9)^2$.

Note also that the distance of $943$ to the true next square is not a square itself, unlike it was for smaller $k$.

I would like to prove that $\forall \, p>2, \, p \in \mathbb P$ the largest $a, \, 0<a<p$ for which the distance to the next square larger than $a p$ is a square itself is $(p-2k)$ where $k$ is the largest such number that the true next square of $p(p-2k)$ is $(p-k)^2$.

As for why I think this would be the case. If $a$ is as described, we get $$a p + k^2 = (p-2k)n+k^2 = p^2 - 2k p + k^2 = (p-k)^2.$$ Thus $$a p = (p-k)^2-k^2.$$ Then note that $$a p = a^2 \left( \left(\frac{p-k}{a} \right)^2 - \left(\frac{k}{a}\right)^2 \right)$$ and so $$\frac{p-k+k}{\gcd(p,a)} = p$$ and $$\frac{p-2 k}{\gcd(p-2 k,a)} = 1.$$ As expected, we get only the two trivial divisors. I have a suspicion that if we could find $a$ which didn't fulfill the condition I wish to prove, it would lead to us discovering nontrivial divisors of $p$, which is impossible. But I would like some ideas or a possible counterexample, than you.

The title of the question comes from the fact that if the distance to next square larger than $p(p-2k)$ is $k^2$ then $(p-k) = \lceil \sqrt{(p-2k) p} \rceil$. The distance to the next largest square is $\lceil \sqrt{(p-2k) p} \rceil^2-(p-2k) p$ and this has to be $k^2$, so $$k^2-(\lceil \sqrt{(p-2k) p} \rceil^2-(p-2k))=(p-k)^2-\lceil \sqrt{(p-2k) p} \rceil^2 = 0$$ and thus $$(p-k)=\lceil \sqrt{(p-2k) p} \rceil$$


1 Answer 1


Find $k$ so that $p-k-1\geq\sqrt{p(p-2k)}$. So $$(p-(k+1))^2 \geq p^2-2k$$ or $$p-2pk -2p + (k+1)^2\geq p^2-2pk$$

Thus $(k+1)^2\geq 2p$. So when $k\geq \sqrt{2p}-1$, you fail to have the given equality, and if $k<\sqrt{2p}-1$ you do.

So the answer to the title question is $k=\lceil \sqrt{2p}-2\rceil$.

You definitely don't need the fact that $p$ is prime, only that it is an integer $p\geq 2$.

  • $\begingroup$ Well, let me demonstrate how this fails if p is not a prime. Let p = 4141 = 41*101. Then 4141*3 + 11^2 = 112^2 and 4141*3 = 112^2-11^2. Then (112+11)/3 = 41. Also 3 = (4141 - 2*2069). But, for example, (4141 - 2*2068) = 5 and the next square of 4141*5 =144^2 and the distance is 31, which is not a square. So what I really wish to prove is that if p is a prime, this kind of thing cannot happen. $\endgroup$
    – Valtteri
    Jul 16, 2014 at 13:11
  • $\begingroup$ But you answered the title question, I think I can use that as well. Thank you. $\endgroup$
    – Valtteri
    Jul 16, 2014 at 13:30
  • $\begingroup$ Wait, if $p=4141$ and then the smallest $k>\sqrt{2p}$ is $k=92$. $\sqrt{p(p-2\cdot 92)}<4141-93$ and $k=91$ does not. $\endgroup$ Jul 16, 2014 at 13:31
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    $\begingroup$ I'm really missing what you are asking, I think, because your comment has a lot of verbiage that seems to have nothing to do with the question I thought I was answering :) $\endgroup$ Jul 16, 2014 at 13:32
  • $\begingroup$ Well, you answered the title question :) I guess I wasn't really clear...The whole question should be more like, if p is a prime and the distance of a*p to the next largest square is itself a square, then the smallest a is of the form (p-2k) and the distance is k^2. In other words, if p is a prime and k is greater than what you gave in your answer, then the distance of p(p-2k) to the next largest square is never a square itself. $\endgroup$
    – Valtteri
    Jul 16, 2014 at 13:39

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