Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

This is most likely going to be my last post before the Bayreuth seminar with the 2nd QED matboj. Expect some nice stolen problems when I am back. $1-\sin ^{5}x-\cos ^{5}x=\left( \sin ^{2}x+\cos ^{2}x\right) -\sin ^{5}x-\cos ^{5}x$ $=\left( \sin ^{2}x-\sin ^{5}x\right) +\left( \cos ^{2}x-\cos ^{5}x\right)$ $=\sin ^{2}x\left( 1-\sin x\right) \left( 1+\sin x+\sin ^{2}x\right) +\cos ^{2}x\left( 1-\cos x\right) \left( 1+\cos x+\cos ^{2}x\right)$ $=\left( 1-\cos ^{2}x\right) \left( 1-\sin x\right) \left( 1+\sin x+\sin ^{2}x\right) +\left( 1-\sin ^{2}x\right) \left( 1-\cos x\right) \left( 1+\cos x+\cos ^{2}x\right)$ $=\left( 1-\cos x\right) \left( 1+\cos x\right) \left( 1-\sin x\right) \left( 1+\sin x+\sin ^{2}x\right)$ $+\left( 1-\sin x\right) \left( 1+\sin x\right) \left( 1-\cos x\right) \left( 1+\cos x+\cos ^{2}x\right)$ $=\left( 1-\cos x\right) \left( 1-\sin x\right) \left( \left( 1+\cos x\right) \left( 1+\sin x+\sin ^{2}x\right) +\left( 1+\sin x\right) \left( 1+\cos x+\cos ^{2}x\right) \right)$ $=\left( 1-\cos x\right) \left( 1-\sin x\right) \left( 2+\left( \sin x+\cos x\right) \left( 2+\sin x+\cos x+\sin x\cos x\right) \right)$. The same factorization is given in the official solution, so I won't bother proving that it can't be improved (anyway, this is not part of the problem). Darij