i) $a^{2}|2^{a}+3^{a}$ and $b^{2}|2^{b}+3^{b}\Longrightarrow (ab)^{2}|2^{ab}+3^{ab}$ ii)5 is 'good' and 55 is 'good' using this 2 facts, we have one solution, because we easily see that the numbers $5^{k}55^{l}$ are 'good's'.

I don't get why i) is true for $a, b$ not being coprime. If it would be true for all $a, b$ you could use just $5$ and you would have $5^{2k}|2^{5^{k}}+3^{5^{k}}$ but that's false. Am I wrong?

TomciO wrote: I don't get why i) is true for $a, b$ not being coprime. If it would be true for all $a, b$ you could use just $5$ and you would have $5^{2k}|2^{5^{k}}+3^{5^{k}}$ but that's false. Am I wrong? You are right $5^{2k}\not |2^{5^{k}}+3^{5^{k}}$ for $k>1$, but $55^{2k}|2^{55^{k}}+3^{55^{k}}$.

So it probably follows from the induction easily (I'm too lazy to check) and we are done, but it doesn't follow from i), I think.

Rust wrote: but $55^{2k}|2^{55^{k}}+3^{55^{k}}$. It seems to me that this is false too. Look, we know that if $\parallel a-b\parallel _{p}=k \geq1$ then $\parallel a^{n}-b^{n}\parallel _{p}=k+\parallel n\parallel _{p}$; but assuming that your clame is correct we should have that $2k \leq \parallel 2^{55^{k}}-(-3)^{55^{k}}\parallel _{5}=\parallel 2+3\parallel _{5}+\parallel 55^{k}\parallel _{5}=1+k$ so $k=1$????But you clamed that $55^{2k}|2^{55^{k}}+3^{55^{k}}$ for all positive integer $k$s,don't you?

We will construct the infinite sequence $ {n_{k}}$ such that for any member $ n_{k}$ of this sequence satisfies the condition: $ 2^{n_{k}} + 3^{n_{k}}$ is divisible by $ n_{k}^{2}$. Easy to verify that $ n_{1} = 1$ and $ n_{2} = 5$. Assume we constructed it until $ n_{k}$. We will easily find $ n_{k + 1}$. Let $ 2^{n_{k}} + 3^{n_{k}} = an_{k}^{2}$,where $ a$ is odd positive integer(as $ n_{k} > = n_{2} = 5$, easy to show that $ 2^{n_{k}} + 3^{n_{k}} > n_{k}^{2}$) and $ a > 1$. Take $ p$ the odd prime divisor of $ a$. Substitute $ 2^{n_{K}}$ by $ x$ and $ 3^{n_{k}}$ by $ y$. Then, obviously ${ x^{p} + y^p} = (x + y)(x^{p - 1} - yx^{p - 2} + ... + y^{p - 1})$. As $ x + y$ is divisible by $ pn_{k}^{2}$, if we define $ r$ to be the reminder of $ x mod p$, than the reminder $ y mod p$ will be $ - r$. So, $ x^{p - 1}, - yx^{p - 2},...,y^{p - 1}$ numbers have the same reminder if we divide them by $ p$, but the number of them is $ p$. So, $ x^{p - 1} - x^{p - 2} + ... + y^{p - 1}$ is divisible by $ p$. It follows $ x^{p} + y^{p}$ is divisible by $ (pn_{k})^{2}$. Now we take $ n_{k + 1} = pn_{k}$. As $ p > 1$, $ n_{k + 1} > n_{k}$ and $ 2^{n_{k + 1}} + 3^{n_{k + 1}}$ is divisible by $ n_{k + 1}^{2}$, which completes our proof. P.S. That's not my own solution. I have read this solution before.

The official solution was(i'm sorry,if it was posted before): If $ n_k$ is the number,such that $ n_k^2|3^{n_k}+2^{n_k}$ then $ n_{k+1}=\frac{3^{n_k}+2^{n_k}}{n_k}$ also satisfies to the given condition. So by taken $ n_1=1$ it follows that $ n_2=5,n_3=55\dots$ and we are done.

Obviously $5$ satisfy $n^2 | 2^n + 3^n$. Now, we'll prove that for each odd number $a$ satisfy, we could construct another larger odd number satisfy. Take an odd prime $q | 2^a + 3^a$ that hasn't been used before. This must exists due to Zsigmondy Theorem. We'll now prove $aq$ satisfy. Now, notice that our hypothesis gives us \[ a^2 | 2^a + 3^a | (2^a)^q + (3^a)^q \]Moreover, $v_q (2^{aq} + 3^{aq}) = v_q (2^a + 3^a) + v_q (q) \ge 2 $. Using this method, one could construct arbitrarily long sequence $a_n$ starting with $a_1 = 5$, and applying the strategy above to find $a_n$ by using $a_{n - 1}$ we're then finished.