Prove that there exist no rational numbers $a,b,c,d$ such that $\left (a+b\sqrt 2 \right )^{100}+\left (c+d\sqrt 2 \right )^{100}=7+5\sqrt 2$
My solutions out line
$\left (a+b\sqrt 2 \right )^{100}+\left (c+d\sqrt 2 \right )^{100}=7+5\sqrt 2$
$\implies\left (a-b\sqrt 2 \right )^{100}+\left (c-d\sqrt 2 \right )^{100}=7-5\sqrt 2$
Which is a contradiction as LHS is positive and RHS is negative.
I hope this will satisfy all of you. I wanted to know any other idea to solve this problem