PP

Number Theory Level 3

Consider all pairs of positive integers x x and y y where exactly one of them is even.

It is possible that ( x + 3 y ) ( 5 x + 7 y ) (x+3y)(5x+7y) is a perfect square?

Yes, it is possible. No, it is not possible.

Áron Bán-Szabó by Áron Bán-Szabó , 17, Hungary

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

note that

( x + 3 y ) ( 5 x + 7 y ) ≡ 1 ( m o d 2 ) It is odd. ( 2 n + 1 ) 2 ≡ 1 ( m o d 8 ) All odd perfect squares are equivalent to 1 ( m o d 8 ) ( x + 3 y ) ( 5 x + 7 y ) = ( x + 3 y ) 2 + 4 ( x + y ) ( x + 3 y ) Both ( x + y ) , ( x + 3 y ) are odd ( x + 3 y ) ( 5 x + 7 y ) ( m o d 8 ) ≡ ( x + 3 y ) 2 ( m o d 8 ) + 4 × ( odd number ) ( m o d 8 ) ≡ 1 ( m o d 8 ) + 4 ( m o d 8 ) ≡ 5 ( m o d 8 ) ⟹ ( x + 3 y ) ( 5 x + 7 y ) cannot be a perfect square when ( x + y ) is odd \begin{aligned}(x+3y)(5x+7y)&\equiv 1\pmod{2}\hspace{5mm}\color{#3D99F6}\text{It is odd.}\\ (2n+1)^2&\equiv 1\pmod{8}\\ \text{All odd perfect squares are}&\text{equivalent to } 1\pmod{8}\\ (x+3y)(5x+7y)&=(x+3y)^2+4(x+y)(x+3y)\\ \text{Both} (x+y),(x+3y)&\text{ are odd}\\ (x+3y)(5x+7y)\pmod{8} &\equiv(x+3y)^2\pmod{8}+4\times(\text{odd number})\pmod{8}\\ &\equiv 1\pmod{8}+4\pmod{8}\\ &\equiv 5\pmod{8} \\ \implies (x+3y)(5x+7y) &\text{cannot be a perfect square when } (x+y) \text{ is odd}\end{aligned}

8 Helpful 0 Interesting 3 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...