Year Sum of Squares

Number Theory Level 3

Let 2017 = a 2 + b 2 2017 = a^2 + b^2 for some positive integers a a and b b .

What is a + b a + b ?


The answer is 53.

Chung Kevin by Chung Kevin , 46, Australia

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.

2 solutions

Viki Zeta
Sep 30, 2016

The equation 2017 = a 2 + b 2 , gives us a circle, with center as (0, 0) \text{The equation } 2017 = a^2 + b^2 \text{, gives us a circle, with center as (0, 0)}

So, there are 4 possible integer solutions,

( 44 , 9 ) , ( − 44 , − 9 ) , ( 9 , 44 ) , ( − 9 , − 44 ) (44, 9), (-44, -9), (9, 44), (-9, -44)

The positive solutions are ( 44 , 9 ) ; ( 9 , 44 ) (44, 9) ; (9, 44) .

∴ a + b = 44 + 9 = 9 + 44 = 53 \therefore a+b = 44+9=9+44=53

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Zee Ell
Sep 30, 2016

4 4 2 + 9 2 = 1936 + 81 = 2017 44^2 + 9^2 = 1936 + 81 = 2017

Hence,

a + b = 44 + 9 = 53 a + b = 44 + 9 = \boxed {53}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Other than checking all cases, how can we know that there is only this solution?

Chung Kevin - 4 years, 8 months ago

Log in to reply

First, by Fermat's Theorem we know that the prime 2017 2017 can be written as the sum of two squares since 2017 ≡ 1 ( m o d 4 ) 2017 \equiv 1 \pmod{4} , so we know we're not on a wild goose chase. The theorem also states that this pairing of squares is unique.

The only possible units digits of a perfect square are 0 , 1 , 4 , 5 , 6 , 9 0,1,4,5,6,9 . To have the sum of perfect squares yield a units digit of 7 7 requires one perfect square with a units digit of 1 1 and one with a units digit of 6 6 . The positive integers n n whose squares end in 1 1 and are less than 2017 2017 are 1 , 9 , 11 , 19 , 21 , 29 , 31 , 39 1,9,11,19,21,29,31,39 and 41 41 . We would then have to go through these cases to find the one for which the difference between 2017 2017 and n 2 n^{2} is a perfect square.

Brian Charlesworth - 4 years, 8 months ago

Log in to reply

Thanks for the response. Alternatively, we can check the other 9 numbers between 1 and 44 (the floor of the square root of 2017) whose squares end in 6: 4, 6, 14, 16, 24, 26, 34, 36 and 44.

Zee Ell - 4 years, 8 months ago

Log in to reply

@Zee Ell – Yes, that would work just as well; each has 9 potential cases to consider. (We would reach the solution faster in my list going from left to right and yours going from right to left. :) )

Brian Charlesworth - 4 years, 8 months ago

Oh. Actually, there is a theorem about the number of ways to express n n as the sum of 2 squares. For primes 4 k + 1 4k+1 , there is only 1 way.

I was reading gaussian integers , but didn't fully understand it.

Chung Kevin - 4 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...