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For positive integer n, let$$ N = 3^{2*3^n} + 3^{3^n} + 1.$$

Two different questions

Is it true that whatever n:

1) the prime decomposition of N contains no prime congruent to 3 modulo 4 raised to an odd power.

2) N is always a square-free integer ?

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    $\begingroup$ What has been tried ? do you know algebra ? $\endgroup$ – user645636 Mar 26 '19 at 19:51
  • $\begingroup$ For the first values of n, in the prime factors of N are all equal to 1 modulo 4 and N is a square free integer. $\endgroup$ – Bernard Vignes Mar 26 '19 at 20:36
  • $\begingroup$ so no known proving techniques ? $\endgroup$ – user645636 Mar 26 '19 at 21:13
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Partial result for a)

$$N=3^{2*3^n} + 3^{3^n} + 1 \equiv ((-1)^{2})^{3^n} + (-1)^{3^n} + 1 \equiv 1 \pmod 4$$

So $N = 1 \pmod 4$. This means that the sum of all powers of primes of form $4k+3$ must be an even number.

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Partial result for (2): suppose that $N$ were not square free and that in fact it is a square. Then $3^{3^n}(3^{3^n}+1)=(x+1)(x-1)$ for some integer $x$. Since $3$ cannot both divide $x+1$ and $x-1$, we conclude that $x=\pm1+k3^{3^n}$ for some integer $k$. In the first case, suppose $x=1+k3^{3^n}$, then $3^{3^n}+1=k(3^{3^n}+2)$, but this is obviously impossible since that would imply $3^{3^n}+2\mid 3^{3^n}+1$, which is absurd. By the same argument it's not possible when $x=-1+k3^{3^n}$ either. So if $N$ is not squarefree, it's at least not a perfect square.

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(1) is equivalent to asking if $N$ can be written as the sum of two squares. And indeed it can $$ N = (3^{3^n}-1)^2 + 3^{3^n+1}. $$

(2) would follow from the fact that $3^{3^n} - 1$ is squarefree for all $n$, since $(3^{3^n}-1)N = 3^{3^{n+1}} - 1$. I'm having a little trouble proving that $3^{3^n}-1$ is always squarefree, though.

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