Proving an absence of rational and integer solutions.

Question

Prove $$(x+y\sqrt2)^2+(z+t\sqrt2)^2=5+4\sqrt2$$ has no solution in rational $(x,y,z,t)$

Prove $$(5+3\sqrt2)^m=(3+5\sqrt2)^n$$ has no solution for positive integers $(m,n)$

How do I approach these kinds of problems? I'm not sure where to start. Also, what are some more problems in this category to practice?

Answer 1

Taking conjugates in $(5+3\sqrt2)^m=(3+5\sqrt2)^n$, we get $(5-3\sqrt2)^m=(3-5\sqrt2)^n$. Now multiply these two equations to obtain $7^m = (-41)^n$, an impossibility.

Using the same technique in $$(x+y\sqrt2)^2+(z+t\sqrt2)^2=5+4\sqrt2$$ we get $$(x-y\sqrt2)^2+(z-t\sqrt2)^2=5-4\sqrt2$$ Multiplying these, we get $$((x+y\sqrt2)^2+(z+t\sqrt2)^2)((x-y\sqrt2)^2+(z-t\sqrt2)^2)= -7 $$ impossible since the left hand side is positive.

Answer 2

Hint for the second part: Take norms: $N(a+b\sqrt2)=a^2-2b^2$. The key property is $N(\alpha\beta)=N(\alpha)N(\beta)$.