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How can I prove (if it's correct) the following:

$$\sin^yt-\cos^yt=1$$ doesn't have solutions such that $y$ is odd and $\tan(t/2)$ is a positive integer.

This is just a small part of a much bigger problem, so I'd like you to check my resolution as well up until here. (Please note that I'm currently not really looking for alternative methods to solve this (for now), unless you believe that proving what I asked above is too complicated.)

I'm supposed to find positive integer solutions for $$(x^2+1)^y-(x^2-1)^y=(2x)^y.$$ I transformed the given equation to $$\left( \frac{2x}{x^2+1} \right)^y + \left( \frac{x^2-1}{x^2+1} \right)^y=1$$ and introduced substitution $$x=\tan{\frac{t}{2}}.$$ Now we can write the equation as $$(\sin t)^2+(-\cos t)^y=1.$$ For $y=1$ we get $x=1$. For $y=2$, every real $x$ (and so every postie integer $x$) is acceptable.

Now for even $y\ge 3$, we get $$\sin^yt+\cos^y<\sin^2y+\cos^2y=1$$, so there are no solutions. But I'm missing what I described in the beginning of the question.

Thanks in advance.

PS. I tried using $x\in(0,3\pi/2)$ because $\tan(t/2)$ has to be positive, but I couldn't get far, even when trying to do each segment on unit circle individually.

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For $y>2$, and not necessarily an integer, there are no solutions except the trivial ones. You have essentially given a proof. – André Nicolas Mar 12 '12 at 22:08
up vote 4 down vote accepted

Since $y \geq 3$

$$ \sin^yt-\cos^yt\leq \sin^2t+\cos^2t=1$$

with equality only if $(\sin(t), cos(t))=(1,0)$ or $(0,-1)$.

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Dang, I totally forgot $-\cos^yt\le \cos^2t$... Too much maths today I guess. Thanks. – Lazar Ljubenović Mar 12 '12 at 22:17

Am I missing something? If you want both $x$ and $y$ to be integers, then you have a restrictive case of Fermat's last theorem so $y$ can only be 0, 1 or 2.

$y=0$ is problematic since only possible $x$ is $x=0$ and who knows what $0^0$ is.

$y=1$ reduces to $2=2x$ or $x=1$.

$y=2$ is the Pythagorean triplets so any $x$ works.

Note: I am not including the trivial case of $x=0$ and $y$ any non-zero even integer.

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