<s>[INST] Is there a simpler way to determine m, n, p, such that the following holds for all reals? I am given the following equation and I am asked to find $m$, $n$ and $p$, so that the equation holds for all reals:
$$ \sin^4x + \cos^4x + m(\sin^6x + \cos^6x) + n(\sin^8x + \cos^8x) + p(\sin^{10}x + \cos^{10}x) = 1, \space \forall x \in \mathbb R $$
I have managed to solve it by, in advance, calculating the following power reduction formulas, and then applying them to the equation:
$$
\sin^4x + \cos^4x = 1 - \frac12\sin^2{2x} \\
\sin^6x + \cos^6x = 1 - \frac34\sin^2{2x} \\
\sin^8x + \cos^8x = 1 - \sin^2{2x} + \frac18\sin^4{2x} \\
\sin^{10}x + \cos^{10}x = 1 - \frac54\sin^2{2x} + \frac5{16}\sin^4{2x}
$$
As a result, I managed to simplify it to the following, which is only in terms of powers of $\sin{2x}$:
$$ \left( 1+m+n+p \right) - \left( \frac12 + \frac{3m}4 + n + \frac{5p}4 \right) \sin^2{2x} + \left( \frac{n}8 + \frac{5p}{16} \right) \sin^4{2x} = 1, \space \forall x \in \mathbb R $$
I then came to the conclusion that the only possible way for which this can be true is if:
$1+m+n+p=1$, $ \frac12 + \frac{3m}4 + n + \frac{5p}4 = 0 $ and $ \frac{n}8 + \frac{5p}{16} = 0 $.
By solving the system of equations below I arrive at the solutions $m=6$, $n=-10$, $p=4$.
$$
\left\{ 
\begin{aligned}
1 + m + n + p = 1 \\ 
\frac12 + \frac{3m}4 + n + \frac{5p}4 = 0 \\ 
\frac{n}8 + \frac{5p}{16} = 0
\end{aligned}
\right. 
$$
My question is if my reasoning is correct and if there exists any simpler way to solve the problem.
 [/INST] Consider the polynomial
$$P(t):=-1+t^2+(1-t)^2+mt^3+m(1-t)^3+nt^4+n(1-t)^4+pt^5+p(1-t)^5\,.$$
If $t=\sin^2(x)$ is a root for every $x\in\mathbb{R}$, then $P\equiv 0$ identically.  In particular,
$$0=[t^4]\,P(t)=2n+5p\,,$$
$$0=[t^2]\,P(t)=2+3m+6n+10p\,,$$
and
$$0=[t^0]\,P(t)=P(0)=m+n+p\,,$$
where $[t^k]\,P(t)$ denotes the coefficient of the term $t^k$ in $P(t)$ for every $k=0,1,2,\ldots$.  (Here, the coefficients are found via the Binomial Theorem.)  It follows immediately that $m=6$, $n=-10$, and $p=4$.
We still need to show that $P\equiv 0$ is indeed the case.  First, $\deg(P)\le4$ can be easily seen.  Therefore, we need to check that $[t]\,P(t)=0$ and $[t^3]\,P(t)=0$, but
$$[t]\,P(t)=-2-3m-4n-5p=-\big([t^2]\,P(t)\big)+\big([t^4]\,P(t)\big)=0$$
and
$$[t^3]\,P(t)=-4n-10p=-2\,\big([t^4]\,P(t)\big)=0\,.$$
Thus, $P$ is indeed the zero polynomial.
</s>