Prove the proposition: $z$ is even if and only if $w$, $x$, and $y$ are even.

Question

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.

Question taken from here.

I am posting this question in order to receive constructive criticism about my proof, and I would also like to improve it and verify it. In addition, I know that this question was already asked here, but I don't need to just better understand the question; I want to come up with a complete and elegant proof for the proposition.

So, I used @ZevChonoles answer from here to help me better understand the question and start constructing my proof.

The first problem I came across when writing this proof is that it is a proposition with if an only if, but it asks to consider all the cases. I use these notes to create my proof templates, but the problem I came across is that the proof template for if an only if propositions and the proof template for case analysis are not exactly the same. I was better able to see how to construct a proof using case analysis, so that is what I went with.

I realized that there could be a total of $4$ different cases, since the order of which numbers are even and which ones are odd does not matter. The cases could be as such:

1) Odd Odd Odd

2) Odd Odd Even

3) Odd Even Even

4) Even Even Even

Theorem: $z$ is even if and only if $w, x,$ and $y$ are even.

Proof: The proof is by case analysis.

None of the numbers $w, x, y$ are odd.
One of the numbers $w, x, y$ is odd.
Two of the numbers $w, x, y$ are odd.
Three of the numbers $w, x, y$ are odd.

Lemma: The square of an odd number is one more than a multiple of $4$:

$$(2j+1)^{ 2 }=4j^{ 2 }+4j+1=4(j^{ 2 }+j)+1$$ and the square of an even number is exactly a multiple of 4:

$$(2i)^{ 2 }=4i^{ 2 }$$

Case 1: If none of the numbers $w, x, y$ are odd, then $z^2=w^2+x^2+y^2=4a+4b+4c$, where $a,b,c\in\mathbb{Z}$, is a multiple of $4$, which implies $z$ is even.

Case 2: If exactly one of the numbers $w, x, y$ is odd, then $z^2=w^2+x^2+y^2=4a+4b+(4c+1)$, where $a,b,c\in\mathbb{Z}$, is one more than a multiple of $4$, and therefore $z$ is odd.

Case 3: If exactly two of the numbers $w, x, y$ are odd, then $z^2=w^2+x^2+y^2=4a+(4b+1)+(4c+1)$, where $a,b,c\in\mathbb{Z}$, is two more than a multiple of $4$, and therefore $z$ is even.

Case 4: If exactly three of the numbers $w, x, y$ are odd, then $z^2=w^2+x^2+y^2=(4a+1)+(4b+1)+(4c+1)$, where $a,b,c\in\mathbb{Z}$, is three more than a multiple of $4$, and therefore $z$ is odd.

Now, from what I understand it seems that $w, x,$ and $y$ being even $\Rightarrow $ $z$ is even, but when I take case $3$ into consideration, it seems that $z$ being even does not necessarily imply that $w, x,$ and $y$ are even.

Am I correct in my conclusion? If not, where did I go wrong in my proof? Please feel free to give me constructive criticism about my proof writing techniques as well.

@shardulc I can tell that you are referring to the mistake I made in Case 3; there is actually no number that is $4k+2$ where $k\in\mathbb{Z}$ that is a square. However, I don't quite understand how that would make $z$ odd. Would you be able to clarify? — Cherry_Developer, Apr 28 '16 at 15:11
Have you see here ? https://en.wikipedia.org/wiki/Pythagorean_quadruple — Emilio Novati, Apr 28 '16 at 15:12
@Cherry_Developer You are right in saying that there is no number whose square is 2 more than a multiple of 4 (can you prove this?). So if two of $w$, $x$, $y$ are odd, then the equation $w^2 + x^2 + y^2 = z^2$ has no solutions at all, so you are done because there is no converse to be proved. — shardulc, Apr 28 '16 at 15:46
@shardulc Alright, so from what I understand, in case $3$), $z^2$ would have to be even, but not a multiple of $4$, but if $z^2$ is even, then $z$ is also even. So, $z^2$ has to be a multiple of $4$, so there is a contradiction that makes the case impossible. What about case $4$? — Cherry_Developer, Apr 28 '16 at 15:54
@Cherry_Developer There is no issue with case 4 because $z$ is not even in that case. The only case in which $z$ exists and is even (case 1) has $w$, $x$, $y$ even, so you have proved the theorem. — shardulc, Apr 28 '16 at 16:00
@shardulc Ok. I think I understand things a bit better now. So, in case 1) $z$ is even and it shows that $z$ is even $\Leftrightarrow w,x,y$ are even is true. Case 2) shows that one of $w, x,$ or $y$ being odd makes $z$ odd. Case 3) gives an impossible case. And finally, case 4) shows that $z$ would be odd. So, case 1) proved what I set out to prove and the other cases did not contradict the proposition. — Cherry_Developer, Apr 28 '16 at 16:08
@shardulc Feel to post your answer below. I will accept it and +1. I'd like to give credit where it's due. I would also like to know if my proof format is good, given that I will apply the proper corrections and a conclusion. — Cherry_Developer, Apr 28 '16 at 16:12

score 1 · Answer 1 · answered Apr 28 '16 at 15:01

1

You are very close. Your lemma tells you what any square looks like: a multiple of 4 or a multiple of 4 plus 1. What does your lemma therefore tell you about case 3?

answered Apr 28 '16 at 15:01

Mees de Vries

26,947

It seems that case $3$ does not fit either case in my lemma. It isn't one more than a multiple of $4$, and it isn't exactly a multiple of $4$. – Cherry_Developer Apr 28 '16 at 15:21
Exactly. You're almost on the right track when you say "how that would make $z$ odd" -- in fact, you already know that $z$ can't be odd, and if it's neither odd nor even then.... – Mees de Vries Apr 28 '16 at 15:25
1

If a number is neither odd nor even, then there is no such number? – Cherry_Developer Apr 28 '16 at 15:35
1

You got it! (Filler word.) – Mees de Vries Apr 28 '16 at 16:59

score 1 · Accepted Answer · answered Apr 28 '16 at 16:19

You are almost right.

In case 1, you have successfully proved that if $w$, $x$, $y$ are all even, then $z$ must also be even. If they are not all even, then $z$ is odd in case 2 and 4. But what about case 3? Here, it can be easily proved that there is no integer whose square is of the form $4k + 2$, hence it is impossible for a solution to exist for case 3.

So out of the four cases, there is only one in which $z$ exists and is even at the same time. That is case 1, and there, $w$, $x$, $y$ are all even; so $z$ being even implies $w$, $x$, $y$ are all even. This concludes your proof.

Note that even though the 'proving' may seem to occur only in case 1 and 3, the other cases are required to cover all the possibilities of $w$, $x$, $y$, otherwise your proof would be incomplete.

score 0 · Answer 3 · answered Apr 28 '16 at 15:25

Yes, you are wrong.

Consider that any perfect square can be written as $3k$ or $3k+1$ where $k \in \mathbb{N}$ since $$x \equiv 0,1,2 \pmod 3 \Rightarrow x^2 \equiv 0,1 \pmod 3$$

Consider now that $3k$ is odd and $3k+1$ is even where $k$ is assumed to be odd. Now proceeding as per the cases mentioned by you, we get that

Case-$1$:$$z^2=3k+3l+3m=3(k+l+m)=3(\text{odd+odd+odd})=\text{odd}$$ Case-$2$:$$z^2=3k+3l+3m+1=3(k+l+m)+1=3(\text{odd+odd+odd})+1=\text{even}$$ Case-$3$:$$z^2=3k+3l+1+3m+1=3(k+l+m)+2=3(\text{odd+odd+odd})+\text{even}=\text{odd}$$ Case-$4$:$$z^2=3k+1+3l+1+3m+1=3(k+l+m+1)=3(\text{odd+odd+odd+odd})=\text{even}$$

But using your condition, you have already shown that Case-$2$ does not hold.

So the answer is out. Hope this helps.

Prove the proposition: $z$ is even if and only if $w$, $x$, and $y$ are even.

3 Answers3