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

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.

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
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 $(0,-1)$.
Dang, I totally forgot $-\cos^yt\le \cos^2t$... Too much maths today I guess. Thanks. –  Lazar Ljubenović Mar 12 '12 at 22:17