Obviosly n is odd. Let $ p$ is maximal prime divisor n, suth that $ n^2|2^n+1$ and $ n=mp$, then $ p|2^m+1$. If $ n=p^km, v_p(m)=0, k\ge 1$, then must be $ v_p(2^m+1)\ge k$. Therefore from all solution n we get $ p|2^1+1$ and find minimal solution (n>1) $ n=3$, Because $ 2^3+1$ had not another prime factors. And all solutions are $ n=1$ and $ n=3$.

Rust wrote: Let $ p$ is maximal prime divisor n, suth that $ n^2|2^n + 1$ and $ n = mp$, then $ p|2^m + 1$. Who says there is such a prime divisor? Rust wrote: If $ n = p^km, v_p(m) = 0, k\ge 1$, then must be $ v_p(2^m + 1)\ge k$. Why?

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Rust wrote: Obviosly n is odd. Let $ p$ is maximal prime divisor n, suth that $ n^2|2^n + 1$ and $ n = mp$, then $ p|2^m + 1$. If $ n = p^km, v_p(m) = 0, k\ge 1$, then must be $ v_p(2^m + 1)\ge k$. Therefore from all solution n we get $ p|2^1 + 1$ and find minimal solution (n>1) $ n = 3$, Because $ 2^3 + 1$ had not another prime factors. And all solutions are $ n = 1$ and $ n = 3$. please write full, I can't understand vp(m) and others... what's mean this function?

For notations, always take a look at http://www.mathlinks.ro/viewtopic.php?t=76610 .

kimnimalar wrote: Rust wrote: Obviosly n is odd. Let $ p$ is maximal prime divisor n, suth that $ n^2|2^n + 1$ and $ n = mp$, then $ p|2^m + 1$. If $ n = p^km, v_p(m) = 0, k\ge 1$, then must be $ v_p(2^m + 1)\ge k$. Therefore from all solution n we get $ p|2^1 + 1$ and find minimal solution (n>1) $ n = 3$, Because $ 2^3 + 1$ had not another prime factors. And all solutions are $ n = 1$ and $ n = 3$. please write full, I can't understand vp(m) and others... what's mean this function? Full solution: Because for $ n>1$ $ 2^n+1$ is odd, n is odd. Let p is maximal prime divisor of n and $ n=mp^k,p\not |m$. Then $ 2^{mp^k}=-1\mod p^{2k}m^2$ and $ (\phi(m^2),p)=1$. If $ m=1$ obviosly $ m^2|2^m+1$. Let $ m>1$.Therefore exist odd u ($ \phi(m^2)$ - even if m>1 and odd), suth that $ p^ku=1\mod \phi(m^2)$. It give $ 2^{mp^ku}=(-1)^u=-1 \mod m^2$. Because $ p^ku=1\mod \phi(m^2)$, we get $ m^2|2^m+1$. It give new solution m. If $ p\not |2^k+1$, then $ 2^{kp}+1=(2^{p-1})^k*2^k+1=2^k+1\mod p$, therefore $ p\not |2^{kp}+1,...p\not |2^{kp^l}$. Because $ p|2^{mp^k}+1$, we get $ p|2^m+1$. Exactly $ p^k|2^m+1$. From solution $ n_0=n=p^km$ we get solution $ n_1=m^2|2^m+1$, suth that $ p|2^m+1$ and $ n_1$ had less prime divisors, then $ n_0$. From $ n_1$ we get $ n_2$,...$ n_l=p^k$ and $ n_{l+1}=1$. Therefore $ p^k|2^1+1=3\to p=3,k=1$. It give solution $ n=3$. If $ n=p^k*3$, then $ p^k|2^3+1=9$. Therefore we had not another solutions.

it's very easy to prove it by the LTE By the LTE we have $v_3(2^n+1)=v_3(2+1)+v_3(n)$ so $v_3(2^n+1)=v_3(3n)$ and as we know $2^n+1$ is divided by $n^2$ so $v_3(2^n+1) \ge v_3(n^2)$ so we replace the LHS by : $v_3(3n) \ge v_3(n^2)$ so we deduce that $n^2$ divides 3n so n divides 3 but n is different to 1 then n must be 3

soundouss wrote: so we replace the LHS by : $v_3(3n) \ge v_3(n^2)$ so we deduce that $n^2$ divides 3n totally wrong!!!

By LTE Vn(2^n+1)=Vn(3) Following n|3 n=1,3 N=3