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Find all natural numbers $n$ such that $2n+1$ and $3n+1$ are square numbers and $2n+9$ is a prime.

I can prove:

$n$ divide by $8$ leaves $0$; $n$ divide by $5$ leaves $0$

So $n$ divide by $40$ leaves $0$ and let $n=40k (k\in N^*)$

And i knew the answer is $n=40\Leftrightarrow k=1$ but i can not how to find $k=1$. Help me, please.

nDLynk
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1 Answers1

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Hint: We have a description of the integers $n$ such that $2n+1$ and $3n+1$ are perfect squares, see

Positive integer $n$ such that $2n+1$ , $3n+1$ are both perfect squares

There are answers referring to Putnam competition where very similar questions have been solved. This, and the above link, might be helpful.

Dietrich Burde
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  • It would help to explicitly state this "complete description" since it it not immediate on perusing the link. In particular, it would be helpful to make it clear how it is of use here. – Bill Dubuque Oct 31 '18 at 17:11
  • @BillDubuque Yes, this would be helpful, I admit. This is only a hint as where to look. I also wanted to refer to the answer by Sandeep Silwal, that $5n+3$ then cannot be prime, etc, which is often asked in Putnam (there are solutions on the web somewhere). I think the case $2n+9$ will go similarly, and there is only $n=40$ possible. – Dietrich Burde Oct 31 '18 at 17:20
  • To clarify, it is not clear to me if you are simply guessing that some result in some answer on that page may be useful here, or if you know for a fact that some specific result there will solve this problem. Could you please elaborate. – Bill Dubuque Oct 31 '18 at 17:24
  • I am guessing (and have some reason) that this has been solved several times in Putnam or other competitions, and that one should give this hint to the OP, so that he will find it. I will edit my answer accordingly. – Dietrich Burde Oct 31 '18 at 17:25
  • Thanks for clarifying. – Bill Dubuque Oct 31 '18 at 17:26