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find all positive integers $n$ for which $n^3+n+1$ is a perfect square

$n^3+n+1$ is odd so $8|n^3+n$ so $8|n(n^2+1)$ if $2|n^2+1$ ,$v_2(n^2+1)=1$ so $n$ must be even, contradiction. So $8\mid n.$

Also by checking $\bmod7$ we understand that $n=0\pmod7$ or $n=2\pmod7$ so $56\mid n$ or $n=16\pmod{56}$ also $72$ is an answer! Another approach: add $n^2$ to both sides: $(n+1)(n^2+1)=a^2+n^2$ It's obvios that $\gcd(a,n)=1$ so by vitek lemma all prime divisors of $a^2+n^2$ are $4k+1$ so the same for prime divisors of $n+1$.

Another approach: let $[\sqrt{n^3}]=c$ So $n^3<(c+1)^2$ so $n+1<c$ so $n^3+n+1<(c+2)^2$ so if $n^3+n+1=a^2$ $a$ is less than or equal to $c+1$ on the other hand it's obvious that $a>c$ so $a=c+1$

Mohammadreza
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The solutions correspond to the integral points on the elliptic curve $$ y^2=x^3+x+1, $$ which are given by $$ (0,\pm 1),\;(72,\pm 611). $$ See for example LMFDB.

It is easy to obtain the points $(72,\pm 611)$ starting from $(0,1)$, see

Find an affine point on $E:y^2=x^3+x+1$

but it requires more on elliptic curves to show that there are no other integer points.

Dietrich Burde
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    It would have been more informative to tell how those solutions were found, and how to show there are no others. – Parcly Taxel May 29 '21 at 14:37
  • Thanks, I hoped an elementary solution exists! I'd be glad if someone presents an elementary solution – Mohammadreza May 29 '21 at 16:33
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    -1 Please see this meta post on "answers" composed purely mechanically. – Bill Dubuque May 29 '21 at 18:54
  • @BillDubuque There are other posts, which have a very similar answer, see for example the answer by Pete Clark here, where the best thing to do is to use a CAS, it seems. Here you did not object so far. I will try to find a better solution, of course. – Dietrich Burde May 31 '21 at 09:04