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I did try to prove this by contradiction(I guess it works). So I suppose this statement is false but I have confused about the meaning of it.

If this statement is false it means "There are odd integers $x,y,z$ according to $(x-z)^2+(y-z)^2=(x+y)^2$." or "There are odd integers $x,y,z$ according to $(x-z)^2+(y-z)^2\neq (x+y)^2$." or.....

Is proof by contradiction a good method? and could you help me how to do next from the statement that is supposed to be false?

Thank you very much.

Bernard
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Mathskid
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  • Pls expand and take out $x^2$ and $y^2$ from both sides. It should be simple from there to show contradiction. – Math Lover Aug 01 '20 at 16:18

2 Answers2

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Hint:

Pls write $x, y, z$ as $2p+1, 2q+1, 2r+1$ respectively. Now expand both sides. You will be able to see the contradiction.

Edit:

You get $2r^2 = 2(pq+qr+pr)+2(p+q)+1$.

Left side is an even number and right side is an odd number.

Math Lover
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2

Write $x, y,$ and $z$ as $2a+1, 2b+1,$ and $2c+1$ respectively.

Then, the equation becomes $(2(a-c))^2+(2(b-c))^2=(2(a+b)+2)^2$, or $4(a-c)^2+4(b-c)^2=(2a+2b+2)^2$.

Expanding both sides, one gets $4(a^2-2ac+c^2)+4(b^2-2bc+c^2)=4a^2+4b^2+4+8ab+8a+8b$.

Dividing both sides by $4$, one gets $a^2-2ac+c^2+b^2-2bc+c^2=a^2+b^2+1+2ab+2a+2b$.

The $a^2$ and $b^2$ terms can then be cancelled, giving $-2ac+c^2-2bc+c^2=1+2ab+2a+2b$.

This gives a contradiction, because the LHS is even, while the RHS is odd.

Geoffrey Trang
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