# Where is the flaw that led to the fallacy that $x^4 + y^4 = z^4$?

Consider $x^2 + y^2 = r^2$. Then take the square of this to give $(x^2 + y^2)^2 = r^4$. Clearly, from this $r^4 \neq x^4 + y^4$.

But consider: let $x=a^2, y = b^2$and$\,\,r = c^2$. Sub this into the first eqn to get $(a^2)^2 + (b^2)^2 = (c^2)^2$. $x = a^2 => a = |x|,$ and similarly for $b.$

Now put this in to give $|x|^4 + |y|^4 = r^4 => (-x)^4 + (-y)^4 = r^4$ or $(x)^4 + (y)^4 = r^4,$ both of which give $x^4 + y^4 = r^4$ Where is the flaw in this argument?

Many thanks.

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Sorry, but I don't understand. What is the statement for which this is a fake proof? – Javier Dec 19 '12 at 17:06
Notice that $x^2+y^2=z^2$ is not always true, it only works for the right choices of $x,y,z$. Notice also that $x^4+y^4=z^4$ is not always false, it does work for some choices of $x,y,z$. – Todd Wilcox Dec 19 '12 at 17:39

$x = a^2$ does not imply that $a = |x|$, rather $|a| = \sqrt{x}$.

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I think there is also a flaw with the statement $(a^2)^2 + (b^2)^2 = (r^2)^2,$ right? Take $a=3, b=4, r=5$ – CAF Dec 19 '12 at 17:02
@CAF: It only assumed that $x^2+y^2=r^2$. Why should we expect that we can have $x+y=r$ (i.e.: $a^2+b^2=c^2$) as well? We need to pick $a,b,c$ appropriately. – Cameron Buie Dec 19 '12 at 20:21

Note that $(x^2+y^2)^2=r^4$ does not imply that $r^4\ne x^4+y^4$.

In fact, you show that $$a^4+b^4=c^4$$ provided $x^2+y^2=r^2, x=a^2, y=b^2, z=c^2$. So what?

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Actually that is wrong too. |Take $a=3, b=4, r=5$ Where is the flaw here? – CAF Dec 19 '12 at 17:21
Why should I be allowed to take $a=3, b=4, r=5$? By your argument, we are restricted to choose $a,b,c$ only so that $x=a^2, y=b^2, r=c^2$ holds for some (assumed given) numbers $x,y,r$ with $x^2+y^2=r^2$. If you want $a=3$, this means that obviously $x=9$. If you want $b=4$,this obviously menas $y=16$. Then from $r^2=x^2+y^2=337$, we can infer, that $r=\pm \sqrt{337}\ne5$. If this is to be the square of a real number $c$, we better take the plus sign and let $c=\pm\sqrt[4]{337}\approx\pm4.2846\ne5$. – Hagen von Eitzen Dec 19 '12 at 17:29

The fact that $x=a^2$ is quite far to imply that $a=|x|$ (second paragraph).

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