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Find all different integers that satisfy the following equality:

$m(\sin^{n}x + \cos^{n} x- 1) = n(\sin^{m}x + \cos^{m}x – 1), (\forall) x\in\mathbb{R}.$

Case1: $m$ is odd, $n$ is even, then put $x=180^0 => m=n=0 =>$ contradiction.

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Case2: $m$ and $n$ is odd, then put $x=180^0 => m=n =>$ contradiction.

Case3: $m$ and $n$ is even $=> m=2a, n=2b$ there $a$, $b$ is integer.

Then put $x=45^0 =>$

$2^{a-b}a(2^{b-1}-1)=b(2^{a-1}-1)$.

Put $x=60^0$ then we have

$4^{a-b}a(4^b-3^b-1)=b(4^a-3^a-1)$. How can I continue?

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In case 3, substitute $u=\sin^2 x$ (and $1-u=\cos^2 x$) to transform the equality into $$a\left(u^b + (1-u)^b – 1\right) = b\left(u^a + (1-u)^a – 1\right)$$

Since the original equality held for all values of $x$, the new one must hold for all values of $u\in \langle 0,1 \rangle$ and thus the polynomials (in variable $u$) on left-hand and right-hand side must be identical.

The degree of the left-hand side is either $b$ (if $b$ is even) or $(b-1)$ (if it is odd, since the highest term cancels out). Similarly, the degree of the right-hand side is either $a$ or $(a-1)$. Without loss of generality, we can assume $a<b$ and since the degrees of the polynomials must be equal, we have $b-1=a$ (and $a$ is even and $b$ odd).

The coefficient at $u^{b-1}$ (which is the same as $u^a$) on the left-hand side is then $(ab)$ and on the right-hand side it’s $(2b)$; equating them yields $a=2$ and $b=3$. It’s easy to check that these values satisfy the equality for all $u$, so the only solution to the original problem is $\{m,n\}=\{4,6\}$.

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