Per the question above I am trying to prove this statement false. As such only one of two conditions have to be met both x and y being odd, or them both being odd. I've seen a lot of examples on this site regarding x^2 - y^2 for a similar example but none where they are added together so I was looking for a pointer on what I might be missing. This is what I have so far: False via contraposition: a^2 + b^2 is not divisible by 8 iff a or b are not even

Assume both a and b are odd A = 2k + 1 and b = 2n + 1 where k and n are both integers

a^2 + b^2 = (2k + 1)^2 + (2n + 1)^2

a^2 + b^2 = 4k^2 + 4k + 1 + 4n^2 + 4n + 1

a^2 + b^2 = 4(k^2 + k + n^2 + n) + 2 (not divisble by 8 so doesnt help the case)

Assume a is odd

A = 2k + 1 and b = 2n

x^2 = 4k(k + 1) + 1 k(k + 1) is even (product of 2 numbers)

x^2 is of form 8k + 1

a^2 + b^2 = 8x + 1 + 8y + 1

a^2 + b^2 = 8(x - y) + 2 (the extra 2 is stoping this case from being true as well)

  • 17
    $\begingroup$ To prove false, a counterexample suffices: $8$ does not divide $2^2+4^2$ $\endgroup$ – J. W. Tanner Apr 15 at 12:06
  • $\begingroup$ @J. W. Tanner Wouldn't we have to provide a counterexample where x^2 + y^2 is divisible by 8 though? $\endgroup$ – Tunifish17 Apr 15 at 12:13
  • $\begingroup$ @Tunifish17 It's only a counterexample if it makes the statement false. Otherwise it's just an example (which is not helpful if you're trying to disprove the statement). $\endgroup$ – Théophile Apr 15 at 12:28

You want to disprove the statement

$$8\mid x^2+y^2\iff 2\mid x, y $$

As @J.W. Tanner pointed out in the comments, a simple counterexample - such as $2^2+4^2$ or its generalization $(2n)^2+(2n+2)^2=8n^2+8n+4\not\equiv 0\pmod 8$ - is enough. This proves that $$2\mid x, y\color{red}{\not\Rightarrow}8\mid x^2+y^2$$

However, you won't be able to disprove the other direction, since it is true $$8\mid x^2+y^2\color{red}{\implies}2\mid x,y$$ This follows from the fact that the only quadratic residues modulo $8$ are $\{0, 1, 4\}$.

  • $\begingroup$ I think I am missing something since doesnt this just prove what we were trying to prove was false? aka that x and y must be even in order to be divisible by 8? $\endgroup$ – Tunifish17 Apr 15 at 13:45
  • $\begingroup$ I fact, $x$ and $y$ have to be even in order $x^2+y^2$ to be divisible by $8$ $\endgroup$ – Dr. Mathva Apr 15 at 14:15
  • $\begingroup$ However, the other direction is wrong, i.e. being even doesn't imply that they will be divisible by $8$ $\endgroup$ – Dr. Mathva Apr 15 at 14:16
  • $\begingroup$ due to the fact that x and y both have to be even in order to be divisible by 8 wouldnt that make the original question true and thus not able to be proved false? $\endgroup$ – Tunifish17 Apr 15 at 15:08
  • $\begingroup$ No, because you said iff. The statement "$8\mid x^2+y^2$" implies that $x,y$ are even is, however, correct. $\endgroup$ – Dr. Mathva Apr 15 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.