L OM - I - Task 5

Find all pairs of positive integers $ x $, $ y $ satisfying the equation $ y^x = x^{50} $.
We write the given equation in the form $ y = x^{50/x} $. Since for every $ x $ being a divisor of $ 50 $, the number on the right side is an integer, we obtain solutions of the equation for $ x \in \{1,2,5,10,25,50\} $. Other solutions of this equation will only be obtained when $ x \geq 2 $ and for some $ k \geq 2 $, the number $ x $ is simultaneously the $ k $-th power of some natural number and a divisor of the number $ 50k $. If $ p $ is a prime divisor of such a number $ x $, then $ p^k|50^k $. Since $ p^k > k $, it cannot be that $ p^k|k $, so $ p \in \{2,5\} $. If $ p = 2 $, then $ 2^k|2k $, from which $ k = 2 $. If, however, $ p = 5 $, then $ 5^k|25k $, from which again $ k = 2 $. Therefore, $ x $ must simultaneously be the square of some natural number and a divisor of the number $ 100 $.
We obtain two new values of $ x $ in this case: $ x = 4 $ and $ x = 100 $. Thus, the given equation has 8 solutions: