$a, b$ are positive whole numbers such that $a + b = a/b + b/a$. How many possible values can $a^4 + b^4$ be? $a, b$ are positive whole numbers such that $a + b = a/b + b/a$. How many possible values can $a^4 + b^4$ be?
I tried using $(a^2 + b^2)^2$ but I don't know what to do after.
 A: $a+b=\frac ab+\frac ba\iff a+b=\frac{a^2+b^2}{ab}\implies a^2b+ab^2=a^2+b^2\implies a^2\underbrace{(b-1)}_{\ge 0}+b^2\underbrace{(a-1)}_{\ge 0}=0$
Since the equation has $a,b$ both on denominator, we can reject zero as an admissible value for $a$ or $b$.
The only possibility for the sum above to be $0$ is that each term is zero, leading to $a=1$ and $b=1$.
A: I propose the following answer, where you can also see a simple way for proving that $a/b+b/a$ is an integer if-f $a=b.$ Following this solution you can also solve the same problem, but with other expressions at the left hand side of the given relation.
Let $d=(a,b).$ Then $a/b+b/a=\frac{a'^2+b'^2}{2a'b'}\in \mathbb{N},$ where $a=da',b=db'$. Obviously $(a',b')=1$ and so, the fact that $\frac{a'^2+b'^2}{2a'b'}\in \mathbb{N}$ implies that
$$a'\mid a'^2+b'^2\Rightarrow a'\mid b'^2\Rightarrow a'=1.$$
Following the same argument with $a'$ and $b'$ interghanged, we get that $b'=1$ too. Consequently, $a=b=d.$ Then $a/b+b/a=2$. Hence, $a+b=2.$ Of course, $a$ and $b$ are $\geq 1,$ as positive integers. Thus, both of them should be equal to $1$. Then the only value of $a^4+b^4$ is $2$.
A: You have $a^2-ba^2+ba-b^2=0$ then $a=\dfrac{-b\pm\sqrt{b^2-4(1-b)(-b^2)}}{2(1-b)}$ so that $5-4b$ must be a square. The only possible value is $b=1$ thus $a=1$ and $a^4+b^4=2$
