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Statement: suppose a,b belongs to Z (integers). If 4/(a^2+b^2) then a and b are not both odd.

By proof of contradiction I assume that a and b are both odd.

If a^2 and b^2 is odd then by definition a and b must be odd too.

It follows that a^2 (or b^2) =(4k+1)^2 <- is this the correct way to show this?

Then a (or b) = 4k+1 <- is this the correct way to show this?

So if a and b are both odd then this is a contradiction hence the supposition is false and the statement is true.

I am wanting to show that a and b are both odd to fit the negation of the statement but I'm unsure about how to show that a and b are both odd in this case?

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Let $4|(a^2+b^2)$, and assume for contradiction that $a,b$ both odd, i.e. :

$a=2k+1$ and $b=2l+1$, for suitable $k,l$.

Then :

$a^2+b^2=(2k+1)^2+(2l+1)^2=4(k^2+l^2+k+l)+2$.


Comments to your answer :

You are assuming that a and b are both odd.

If $a^2$ and $b^2$ is odd then by definition a and b must be odd too.

It is the other way round : if we assume that a and b are both odd, then $a^2$ and $b^2$ are odd too.

Assume $a,b$ both odd; i.e. $a=2k+1$ and $b=2l+1$. Then $a^2=(2k+1)^2=4k^2+2k+1$ and the same for $b^2=(2l+1)^2$.

But why : $a^2 = (4k+1)^2$ ?

It is not true that every odd number is of the form $4k+1$; try with $a=7$.

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  • $\begingroup$ Could you please clarify your last step: 'But then : a^2=4m by divisibilty by 4 ..' I don't quite understand how you got a^2 = 4m? $\endgroup$ – joe Sep 21 '14 at 8:25

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