Prove that $(3+5\sqrt{2})^m=(5+3 \sqrt{2})^n$ has no positive integer solutions?

Question

Is my proof ok? I set $b=3+5 \sqrt{2}$, so that we have $b^m=(b+2-3 \sqrt{2})^n$ , or $b^m=(b+\sqrt2(\sqrt2- 3))^n$. Since $RHS<LHS$, $n>m$ . However, from what we know about binomial expansion, we will have a $b^n$ on the $RHS$ which will not cancel out with anything since it is the highest power, so we will never be able to reduce the $RHS$ to just $b^m$ . I don't really like my proof, is there a better one? And does this have a solution for rational powers, and how can you prove that it does/doesn't? Thanks.

Answer 1

If there is a solution, there is one where the $\sqrt{2}$are replaced by $-\sqrt{2}$. Multiply. We get $$(3+5\sqrt{2})^m(3-5\sqrt{2})^m=(5+3\sqrt{2})^n(5-3\sqrt{2})^n.$$ That gives $(-41)^m=7^n$, impossible unless $m=n=0$.

Can't have non-zero rationals either. Just raise to the appropriate integer power and use basically the same argument.

Answer 2

I don't understand your statement about not cancelling.

Hint: consider the function $f(a + b \sqrt{2}) = (a + b \sqrt{2})(a - b \sqrt{2}) = a^2 - 2 b^2$, and note that $f(xy) = f(x) f(y)$.