Number Theory Problem

Well, first of all we note that a divides $c^{25}$: if b divides $c^5$, then $b^5$ divides $c^{25}$. Because a divides $b^5$, it divides also $c^{25}$. This can be applied cyclically: b divides $a^{25}$ and c divides $b^{25}$.

In the factorization of $(a+b+c+)^{31}$ there will be three members $a^{31};b^{31};c^{31}$, members like $n\cdot a^k \cdot b^{31-k}$ (and cyclical, i.e. others with ac or bc instead of ab) and members like $abc \cdot something$. Clearly, $abc$ divides all the members of the third type. We can express $a^{31}$ as $a^{25}\cdot a^5 \cdot a$, for the observation written above, it's dividible by $abc$.

Remains the second case:let's assume WLOG that's with the letters a and b.

We have three cases: k<5,k=5 and k>5.

In the first one, $31-k\geq 26$, so we can express $a^k\cdot b^{31-k}$ as $a^k cdot b^{25}\cdot b^h$ where h is a positive integer. a divides $a^k$, b divides $b^h$ and c divides $b^25$ so abc divides $a^k\cdot b^{31-k}$ .

In the second case,$a^k\cdot b^{31-k}$ is $a^5\cdot b^{21}\cdot b^5$. c divides $a^5$, b divides $b^{21}$ and a divides $b^5$.

In the third case, we can express $a^k\cdot b^{31-k}$ as $a^5 cdot a^j \cdot b^{31-k}$ where j is a positive integer. c divides $a^5$, b divides $b^{31-k}$ and a divides $a^j$. Q.E.D.

A better solution would be choose any prime $p$ dividing $a,b,c$ and consider the maximum powers of $p$ in $a,b,c$ respectively $\alpha,\beta,\gamma$ and WLOG let $\alpha\ge \beta \ge \gamma$ .Now the max power of $p$ dividing $(a+b+c)^{31}$ is $p^{31\gamma}$ .It suffices to show that $\alpha+\beta+\gamma\le 31\gamma$ but $\alpha\le 5\beta\le 25\gamma$ from the given condition.So summing them we get the desired inequality.This helps an easy generalization: let $a,b,c$ be natural numbers such that $a\mid b^n,b\mid c^n,c\mid a^n$ then $abc\mid (a+b+c)^{n^2+n+1}$