An Odd Sum

Number Theory Level 2

3 distinct integers satisfy

$a^2 + b^2$ is odd
$a^2 + c^2$ is odd

What can we say about $b^2 + c^2$ ?

It must be even It must be odd It can be odd and can be even

1 solution

Brian Charlesworth
Oct 27, 2016

Since the sum of two odd integers is even, we know that $(a^{2} + b^{2}) + (a^{2} + c^{2}) = 2a^{2} + (b^{2} + c^{2})$ is some even number $d$ . Then $b^{2} + c^{2} = d - 2a^{2}$ is $\boxed{even}$ since the difference between two even numbers, in this case $d$ and $2a^{2}$ , is always even.

@Chung Kevin Sorry about the report, (now deleted), to your other question. For some brain-fart reason I thought we were to count the number of pairs $(n,m)$ . :p

Brian Charlesworth - 4 years, 7 months ago

Ah. When I wrote the question, I first asked for pairs $(n,m)$ but then it felt pointless to do so once you see the equation, so I settled just for $t$ .

Chung Kevin - 4 years, 7 months ago

We actually didn't need that these terms are squares, just that they are integers.

I'm was looking at a version where squares are involved, and also conditioned to primes, but it ended up being confusing / not that nice.

Chung Kevin - 4 years, 7 months ago

An Odd Sum

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