Finding a root of a degree 5 polynomial
Solution 1:
Your second idea is better $$(a+b)^5 = a^5+b^5 + 5ab(a^3+b^3)+10a^2b^2(a+b)$$
So $$x^5 = 2m + 10(a+b)\Big((a+b)^2-6\Big)+40(a+b)$$ or $$x^5+2m+10x^3-60x+40x$$ so $$x^5-10x^3+20x -2m =0$$ and thus $m=20$ so ...