Problem 839

Noticing that $30621=173 \cdot 177$, and introducing the short notations $177=a, 173=b$, our numbers are

$$\begin{aligned} & A=a^{5}+a b^{4}-b^{5} \\ & B=b^{5}+a^{4} b-a^{5} \\ & C=a^{4}+a^{2} b^{2}+b^{4} \end{aligned}$$

Accordingly, $B$ results from $A$ by swapping $a$ and $b$. This swap maps $C$ onto itself, so it suffices to examine the question of $A$ and $C$.

$C$ can be transformed as follows:

$$\left(a^{2}+b^{2}\right)^{2}-a^{2} b^{2}=\left(a^{2}+b^{2}-a b\right)\left(a^{2}+b^{2}+a b\right)$$

Based on this, we will separately consider the question of the greatest common divisor with $A$ for each of the two factors.

A natural question is whether there exist $M, N$, coefficients independent of $a$ and $b$, such that the polynomial

$$a^{3}+M a^{2} b+N a b^{2}-b^{3}$$

of $a$ and $b$, when multiplied by the first factor of $C$, gives $A$. This is satisfied if the coefficients of the like terms in $A$ and the product are respectively equal (which we have already ensured for the coefficients of $a^{5}$ and $b^{5}$ by choosing the coefficients of $a^{3}$ and $b^{3}$), that is,

$$\begin{array}{lll} a^{4} b & \text { from the coefficients } & 0=M-1 \\ a^{3} b^{2} & \text { from the coefficients } & 0=N-M+1 \\ a^{2} b^{3} & \text { from the coefficients } & 0=-1-N+M \\ a b^{4} & \text { from the coefficients } & 1=1+N \end{array}$$

It is easy to see that there is a solution: $M=1, N=0$, so $A$ is divisible by $a^{2}-a b+b^{2}$, the quotient being $a^{3}+a^{2} b-b^{3}$; further, we seek the common divisor of this and the second factor of $C(\mathbf{1})$.

According to a known identity, the quotient

$$\left(a^{3}-b^{3}\right)+a^{2} b=(a-b)\left(a^{2}+a b+b^{2}\right)+a^{2} b$$

Accordingly, if there exists the common divisor mentioned now, it is also a divisor of the last term on the right side, $a^{2} b$. The divisors of this (since in our problem $a$ and $b$ are relatively prime, as $b$ is prime and not a divisor of $a$) are the divisors of $b$ and $a^{2}$. However, $b$ has no common divisor greater than 1 with $a^{2}+a b+b^{2}$, because $a$ is a divisor of $b$ and $b^{2}$, but $a^{2}$ is not divisible by any divisor of $b$ greater than 1, and $a^{2}$ has no common divisor greater than 1 with $a^{2}+a b+b^{2}$, because the sum of the last two terms is $b(a+b)$, where $b$ is prime and $a+b$ is relatively prime to $b$, to $a$, and thus to $a^{2}$.

Accordingly, the greatest common divisor of $A$ and $C$ is $a^{2}-a b+b^{2}=177^{2}-30621+173^{2}=30637$, and this is also the greatest common divisor of the pair $B, C$ based on the aforementioned (swap) relationship between $A$ and $B$, since swapping $a$ and $b$ does not change $a^{2}+a b+b^{2}$, and the reasoning applied to $a^{2} b$ can essentially be repeated for $b^{2} a$.