# Pythagoras theorem quick calculation [closed]

How can I easily calculate this equation $x^2+y^2=76149513$ when $x$ and $y$ are whole numbers?

## closed as off-topic by Namaste, Siong Thye Goh, Simply Beautiful Art, Arnaud Mortier, user99914 Aug 1 '18 at 16:44

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• Are you asking for pairs of solutions $x$ and $y$? – Matt Aug 1 '18 at 12:47
• Yes and also a shortcut way for his sort of caculations – Mathisfun Aug 1 '18 at 12:58

Wolfram alpha tells me $$76149513 = 3^2×11×353×2179 .$$
Since $11$ is a prime factor congruent to $3$ modulo $4$ that number can't be written as a sum of two squares.
(I could have tested for divisibility by $11$ by calculating the alternating sum of the digits.)
• Note that divisibility by $11$ alone is not enough to show that the number is not a sum of two squares; you also need the odd power here. – Dirk Aug 1 '18 at 13:20
• @DirkLiebhold True. It's enough when $11$ occurs to an odd power - and easy to prove when the power is $1$. – Ethan Bolker Aug 1 '18 at 13:59