An algebra problem by Sean Ty

Firstly, we prove that $\sqrt{7}-\sqrt{5}<1$ .

Proof: Since $121<140$ , we have that $11<2\sqrt{35}$ or $12-2\sqrt{35}<1$ . Taking the square root of both sides gives us the desired $\sqrt{7}-\sqrt{5}<1$ .

Now consider the expression $(\sqrt{7}+\sqrt{5})^6+(\sqrt{7}-\sqrt{5})^6$ . When we expand it using the binomial theorem, all terms with an odd power of $\sqrt{5}$ (and thus all odd powers of $\sqrt{7}$ ) all cancel, leaving, only an integer to deal with. Expanding gives

$(\binom{6}{0}\sqrt{7}^6+\binom{6}{1}\sqrt{7}^5(\sqrt{5})^1+\cdots+\binom{6}{6}(\sqrt{5}^6)-(\binom{6}{6}\sqrt{7}^6+\binom{6}{1}\sqrt{7}^5(-\sqrt{5})^1+\cdots+\binom{6}{6}(-\sqrt{5})^6)$ $=2(7^3)+2\binom{6}{2}(7^2)(5)+2\binom{6}{4}(7)(5^2)+2(5^3)$ $=13536$ .

Earlier we proved that $\sqrt{7}-\sqrt{5}<1$ so that $(\sqrt{7}-\sqrt{5})^6<1$ . Since $(\sqrt{7}+\sqrt{5})^6+(\sqrt{7}-\sqrt{5})^6=13536$ , then we have that $(\sqrt{7}+\sqrt{5})^6=13536-(\sqrt{7}-\sqrt{5})^6$ . Thus $13536-1<(\sqrt{7}+\sqrt{5})^6<13536$ so $\lfloor{(\sqrt{7}+\sqrt{5})^6}\rfloor=\boxed{13535}$ .

نظرية ذات الحدين:D