# Prime decomposition

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 ?

• What has been tried ? do you know algebra ? – user645636 Mar 26 '19 at 19:51
• 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. – Bernard Vignes Mar 26 '19 at 20:36
• so no known proving techniques ? – user645636 Mar 26 '19 at 21:13

$$N=3^{2*3^n} + 3^{3^n} + 1 \equiv ((-1)^{2})^{3^n} + (-1)^{3^n} + 1 \equiv 1 \pmod 4$$
So $$N = 1 \pmod 4$$. This means that the sum of all powers of primes of form $$4k+3$$ must be an even number.
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.
(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.