mathisreal wrote: Let $m$ be a positive integer. Find the number of real solutions of the equation $$|\sum_{k=0}^{m} \binom{2m}{2k}x^k|=|x-1|^m$$ Note that $x=0$ is never a solution (since then $x^k$ is not defined in LHS when $k=0$) Note that $m=1$ implies no solution (since $x=0$ is never a solution) If $m\ge 2$ : 1) $x>0$ Let $x=y^2$. Equation is $\frac{(1+y)^{2m}+(1-y)^{2m}}2=2|1+y|^m|1-y|^m$ Which implies $((1+y)^m\pm(1-y)^m)^2=0$ Which has unique solution $y=0$ And so $x=0$, which is not a positive solution. 2) $x<0$ Let $x=-y^2$ for some $y>0$ Equation is $\left|\frac{(1+iy)^{2m}+(1-iy)^{2m}}2\right|=2(1+iy)^m(1-iy)^m$ Which is $((1+iy)^m\pm(1-iy)^m)^2=0$ and so $(1+iy)^{2m}=(1-iy)^{2m}$ And so (after some rather simple computations) $y=\tan \frac{k\pi}{2m}$ whatever is $k\in\{1,2,...,m-1\}$ And so $\boxed{\text{Exactly }m-1\text{ solutions : }x=-\tan^2 \frac{k\pi}{2m}\quad\forall k\in\{1,2,...,m-1\}}$