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This is a problem from this year´s local Olympiad exam from Catalunya (2022/2023).

I have tried solving this problem without success. The only useful thing I came up with is that it might be solved by inequalities as its intervals are positive integers and not real numbers. I think I must have forgotten some important concepts that I would be grateful if you could tell me which.

I wish you could help me solve this problem. Thank you very much!!

Still_Learning
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    Can you factors the quadratic in $n$? – Mark Bennet Dec 25 '22 at 17:23
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    Similar questions: https://math.stackexchange.com/q/830009/42969, https://math.stackexchange.com/q/1128670/42969, https://math.stackexchange.com/q/2272982/42969 – Martin R Dec 25 '22 at 17:25
  • You mean in (n+2)(n+3)? – Still_Learning Dec 25 '22 at 17:25
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    or $ (n+2)^2 < n^2 + 5n + 6 < (n+3)^2$ which would still work for altered polynomial, $ (n+2)^2 < n^2 + 5n + 7 < (n+3)^2$ where this one does not factor – Will Jagy Dec 25 '22 at 17:25
  • Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. – Martin R Dec 25 '22 at 17:27
  • Is possible to do that Will? Is it any formula? – Still_Learning Dec 25 '22 at 17:27
  • $N=(n+2)(n+3)$ so $n+2=ax^2,\space\space n+3=by^2$ with $ab=$ square. It seems that solving the equation in $n$ is easier and inmediate. because the discriminant equal to $1$. – Piquito Dec 25 '22 at 17:34
  • Hi Martin. The problem is only the title, it does not have extra information. The point that you are trying to emphasize is that I have to put the question again in the body? – Still_Learning Dec 25 '22 at 17:36
  • Hi @Piquito. Thanks for your help. But there is one understand, how did you know that n+2=ax2, n+3=by2 ?? – Still_Learning Dec 25 '22 at 17:41
  • I am quite a beginer and I am trying to learn. Sorry if the question is elementary. – Still_Learning Dec 25 '22 at 17:42
  • Alternate approach: $$n^2 + 5n + 6 = k^2 \implies 4n^2 + 20n + 24 = 4\left(k^2\right) \implies $$ $$(2n+5)^2 - 1 = (2k)^2 \implies $$ $$[(2n+5) - 2k)] \times [(2n+5) + 2k] = 1.$$ – user2661923 Dec 25 '22 at 20:58
  • Hi @Still_learning. I saw that $n+2$ and $n+3$ are consecutive but anyway my first assertion was deficient and the second one it would be preferable. In any time it was been proved that the product of consecutive integers is never a power. – Piquito Dec 26 '22 at 17:20
  • Thanks! But what about 0 and 1? That product is a perfect square. – Still_Learning Dec 26 '22 at 17:32

2 Answers2

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There are no integer solutions because

$$(n+2)^2 = n^2+4n+4 \lt n^2+5n+6 < n^2 + 6n + 9 = (n+3)^2$$

and there are no other integers between $n+2$ and $n+3$.

yagod
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Alternate solution: we solve over the integers.

$n^2 + 5n + 6 = (n + 2)(n + 3)$. Now $n + 2$ and $n + 3$ are comprime, so if their product is a square, both of these factors must also be squares (or both must be the negation of squares). The only squares 1 apart are $0$ and $1$, so we must have $n = -2$ or $n = -3$. These are the only solutions over the integers, and neither is positive.

Mark Saving
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