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2025 SSMO Team Round Problems/Problem 2

Problem

Nonnegative integers $a$ and $b$ satisfy $a!+b^4 = 16a+41$. Find the sum of all possible values of $b$.

Solution

Rearrange the given equation as $b^4 = 16a+41-a!$. The left hand side is always nonnegative, but the right hand side is only nonnegative when $a\le 5$. Therefore, we need only check six possibilities for $a$.

  • If $a = 0$, then $b^4 = 40$, which is impossible.
  • If $a=1$, then $b^4 = 56$, which is impossible.
  • If $a=2$, then $b^4 = 71$, which is impossible.
  • If $a=3$, then $b^4 = 83$, which is impossible.
  • If $a=4$, then $b^4 = 81$, so $(4,3)$ is a solution.
  • If $a=5$, then $b^4 = 1$, so $(5,1)$ is a solution.

The sum of all possible values of $b$ is $3+1 = \boxed{4}$.

~Sedro

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