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Problem:

Suppose that $w^2+x^2+y^2=z^2$, where $w$, $x$, $y$, and $z$ always denote positive integers. (Hint: It may be helpful to represent even integers as $2i$ and odd integers as $2j+1$, where $i$ and $j$ are integers)

Prove the proposition: $z$ is even if and only if $w$, $x$, and $y$ are even. Do this by considering all the cases of $w$, $x$, $y$ being odd or even.

The accepted solution:

In fact, it is easy to see that the converse holds as well. That is, for any integer $m$,

$m$ is even $\iff m^2$ is a multiple of 4,

$m$ is odd $\iff m^2$ is one more than a multiple of 4.

My understanding:

If $m$ is even, then $m^2$ is a multiple of 4, does not always mean: if $m^2$ is a multiple of 4, then $m$ is even.

With $w = 2$, $x = 4$, $z = 6$:

$$z^2=2^2+4^2+6^2=4+16+36=56$$

$z=\sqrt{56}\approx 7.48$ which is not an even integer. I would like to understand the accepted solution.

bb_823
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chenel
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    The accepted solution implicitly says "Assuming we have a solution quadruple of positive integers such that ....", which contradicts your "counterexample" because $z$ is not a positive integer. – Calvin Lin Jun 16 '23 at 16:13
  • Notice your quoted piece of the solution says "for any integer $m$". So the claim $m^2$ is a multiple of 4 $\Rightarrow$ $m$ is even is talking about cases where $m$ is an integer and $m^2$ is a multiple of 4, but says nothing at all about cases where $m^2$ is a multiple of 4 but $m$ isn't actually an integer. – aschepler Jun 16 '23 at 16:21

1 Answers1

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In the problem the following is stated: "where $w$, $x$, $y$, and $z$ always denote positive integers". This means that you should be ignoring all cases where $z$ isn't a positive integer, such as the example you've given.

bb_823
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