10. Let $a, b$ be positive integers that are coprime. What is the maximum value that the greatest common divisor of $(a+b)^{4}$ and $a-b$ can take?
(A) 3
(B) 4
(C) 16
(D) 32
(E) can be arbitrarily large.

## Numerical answer problems - 8 points
10. The answer is (C). First, consider $d=\text{GCD}(a+b, a-b)$. The GCD of two numbers also divides their sum and difference, so $d$ divides $2a$ and $2b$. Since $a$ and $b$ are coprime, no odd prime can appear in $d$, and the prime 2 can appear with an exponent of at most 1 (this happens if and only if $a$ and $b$ are both odd).

For the case $c=\text{GCD}\left((a+b)^{n}, a-b\right)$, observe that the prime factors of $(a+b)^{n}$ are the same as those of $a+b$, with the exponents multiplied by $n$, so $c$ is a power of 2 (and remains 1 if $a$ and $b$ are not both odd). However, $a+b$ and $a-b$ cannot both be multiples of 4, because their difference $2b$ is not.

Now, if $a-b$ is not a multiple of 4, then $c$ cannot be either. Thus, the only remaining case is when $a+b$ is not a multiple of 4, and in this case $c$ can be at most $2^{n}$. This value can always be achieved, for example by taking $a=2^{n}+1, b=1$.