Problem 1

Find all quadruples of positive integers

(a, p, m, o)

with

o \ge 2

such that

a! + p! = m^o + 26

Step 5 of 5: Assemble all quadruples

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(a,p,m,o)\in\{(4,3,2,2),(3,4,2,2),(5,3,10,2),(3,5,10,2)\}

Detailed analysis

Combining the three cases, the only solutions with $a\ge p$ are $(a,p,m,o)=(4,3,2,2)$ and $(5,3,10,2)$ . Since the equation is symmetric in $a$ and $p$ , swapping the two coordinates gives $(3,4,2,2)$ and $(3,5,10,2)$ . These four quadruples are exactly the solutions.