# Show that if n is an even perfect number then n is not the sum of two squares. [closed]

• A perfect number is a positive integer that is equal to the sum of its proper divisors, and all perfect numbers are even.

• A sum of two squares is an integer that is the sum of two squares integers.

• The body of your question is unnecessary. People here know what perfect numbers and sums of squares are. And while the only perfect numbers known are even, the existence or non-existence of odd perfect numbers is open (unsolved). Rather a famous open problem, in fact. Finally, you need to give more context. Where is this problem from? What are your thoughts? Apr 6, 2020 at 4:21
• Do you know the factorization of perfect numbers? Do you know that a number of the form $4n+3$ is not the sum of two squares? Apr 6, 2020 at 4:29

By a theorem of Euler, all even perfect numbers are of the form $$2^{n-1}(2^n-1)$$, where $$2^n-1$$ is a (Mersenne) prime. But $$2^n-1\equiv3\mod4$$ for all perfect numbers ($$n\ge2$$), and it appears only once in the prime factorisation, so by the sum of two squares theorem, the perfect number is not expressible as the sum of two squares.