Lemma: $p|3^x-2^x$ and $p|3^y-2^y$ with $x$ is minimal then $x|y$ Proof: Let $y=xk+r$ with $0<r<x$ Then $p|3^{xk+r}-2^{xk+r}$ $(1)$ But $p|3^x-2^x \Rightarrow p|3^{xk}-2^{xk} \Rightarrow p|3^{xk+r}-2^{xk}.3^r$ $(2)$ From $(1)(2)$ we gain $p|2^{xk}.3^r-2^{xk+r} \Rightarrow p|2^{xk}(3^r-2^r) \Rightarrow p|3^r-2^r$ But $r<x$ so this gives contradiction because $x$ is minimal Let $p$ be the least prime divisor of $n$ Then $p|3^n+2^n \Rightarrow p|3^{2n}-2^{2n}$ Let $k$ be the smallest number such that $p|3^{k}-2^k$ Then by lemma we gain $k|2n$ but by FLT we have $k|p-1$ Thus $gcd(k,n)=1$ because if $r|k$ and $r|n$ with $r$ is a prime then $r|p-1 \Rightarrow r<p$ and $r$ is the prime divisor of $n$ this contrary to the minimal of $p$ Then $k|2 \Rightarrow k=1,2$ If $k=1 \Rightarrow p|1$ false! If $k=2 \Rightarrow p=5$ Then $n=5^x.y$ Thus $5^{2x}|3^{5^x.y}+2^{5^x.y}$ But by LTE $v_5(3^{5^x.y}+2^{5^x.y})=v_5(2+3)+v_5(5^x.y)=1+x$ Thus $1+x \geq 2x \Rightarrow x=1$ Then $n=5.y$ with $gcd(5,y)=1$ and now $25y^2|3^{5y}+2^{5y}$ then $y^2|(3^5)^y+(2^5)^y$ This time we similarly suppose $p'$ is the least prime divisor of $y$ then similarly $k=2$ then $p'|(3^5)^2-(2^5)^2$ thus $p'|3^5+2^5$ (because if $p|3^5-2^5$ this give contradiction because $2$ is the smallest number such that $p'|(3^5)^k-(2^5)^k$) and this time we will choose similarly $n=5.p'.y'$ with $gcd(y',p)=gcd(y',5)=1$ then by LTE we can get $p'^2|(3^5)^{p'y'}+(2^5)^{p'y'}$ Thus we have $y'^2|(3^{5.p'})^{y'}+(2^{5.p'})^{y'}$ The process will continue and we gain there exists infinite $n$ satisfying the problem (Q.E.D)

This was Problem 1. of Paolo Leonetti's (bboypa) Oliforum contest 2012. We will start by proving a couple of lemmata. Lemma 1. If an odd prime $p$ divides $a+b$, then $p^2$ divides $a^p + b^p$. Proof. One can invoke the Lifting The Exponent lemma, but a direct proof is straightforward enough. We have \[a^p+b^p = (a+b)(a^{p-1} - a^{p-2}b + \cdots - ab^{p-2} + b^{p-1}).\] But $p\mid a+b$ means $b\equiv -a \pmod{p}$, and so \[a^{p-1} - a^{p-2}b + \cdots - ab^{p-2} + b^{p-1} \equiv pa^{p-1} \equiv 0 \pmod{p}.\] Therefore $p^2 \mid a^p + b^p$. Lemma 2. If a prime $q\neq 5$ divides $2^p+3^p$, with $p$ prime, then $2p<q$. Proof. Again, this probably can be argued by {\sc Zsigmondy} theorem, but we will provide a short direct proof. We have $2^p\equiv -3^p \pmod{q}$, so $(2\cdot 3^{-1})^{p} \equiv -1 \pmod{q}$. Let $\nu$ be the (multiplicative) order of $2\cdot 3^{-1}$ modulo $q$, so $\nu \mid 2p$ and $\nu \mid q-1$, hence $\nu \leq q-1$. But clearly $\nu > p$ (since $\nu = 1$ leads to $2\equiv 3 \pmod{q}$, $\nu = 2$ leads to $4\equiv 9 \pmod{q}$, while $\nu = p$ leads to $2^p \equiv 3^p \pmod{q}$, all absurd). Therefore $2p = \nu \leq q-1$, hence $2p<q$. Solution. Take now $p_1 = 5$, and $p_k \neq 5$ a prime divisor of $2^{p_{k-1}} + 3^{p_{k-1}}$ for $k\geq 2$. We can then only take $p_2 = 11$, since $2^{p_{1}} + 3^{p_{1}} = 2^5 + 3^5 = 5^2\cdot 11$. Notice now that for any odd prime $p\neq 5$ we have \[2^p + 3^p = 5(2^{p-1} - 2^{p-2}\cdot 3 + \cdots - 2\cdot 3^{p-2} + 3^{p-1}) \equiv 5p2^{p-1} \not \equiv 0 \pmod{5^2},\] therefore $2^p + 3^p$ has a prime divisor different from $5$. By Lemma 2 we will have $p_1<p_2<\cdots < p_m < \cdots$ (the next one, $p_3=35839$, is quite large!). Let $n_m = p_1p_2\cdots p_m$. For $k = 1$ we have $p_1^2 = 5^2 \mid 2^{5} + 3^{5} \mid 2^{n_m} + 3^{n_m}$, while for $2\leq k\leq m$ we have $p_k \mid 2^{p_{k-1}} + 3^{p_{k-1}}$, hence $p_k^2 \mid 2^{p_{k-1}p_k} + 3^{p_{k-1}p_k} \mid 2^{n_m} + 3^{n_m}$, by Lemma 1, so $n_m^2 \mid 2^{n_m} + 3^{n_m}$.

Yes, your solution is correct, and in my solution, $p_i|3^{p_{i-1}}+2^{p_{i-1}}$ then by my lemma in my proof, $p_i \vdots 2p_{i-1}$ so $p_i>p_{i-1}$ then the process will continue

mavropnevma wrote: This was (also) Problem 1. of Paolo Leonetti's (bboypa) Oliforum contest 2012. Hi Dan, you can remove the "also"