Taking a look at this year
Is 2017 expressible as the sum of two perfect squares ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Sir how did you generate the numbers 9 and 44 and thank you for the solution as it is the best way to check whether any Prime can represented as the sum of two squares
Log in to reply
Well, there is some brute force involved. First we note that the last digits of perfect squares, (i.e., quadratic residues modulo 10), are 0 , 1 , 4 , 5 , 6 , 9 . Thus since we require a sum that ends in 7 we must have one square that ends in 1 and one that ends in 6 . Now each square must be less than 2 0 1 7 , so the only integers we can use whose squares end in 1 are 1 , 9 , 1 1 , 1 9 , 2 1 , 2 9 , 3 1 , 3 9 and 4 1 . So since this is a short list we just need to subtract the square of each of these numbers from 2 0 1 7 and see if we end up with another perfect square. Doing this, we find that 2 0 1 7 = 9 2 + 4 4 2 is the only option.
(We could have looked at all the squares that end in 6 also but this is a longer list so would have taken more time.)
By Fermat's 4n + 1 Theorem , since 2 0 1 7 is prime and 2 0 1 7 ≡ 1 ( m o d 4 ) we are guaranteed that it can be expressed as the sum of two perfect squares. So the answer is Yes .
Specifically, 2 0 1 7 = 9 2 + 4 4 2 , (which is in fact the only solution for positive x , y .