If f = (x/y) + (y/x) +1/(xy) is an integer. Prove that f must be of the form 3x

Question

I have tried using Induction method but I am unable to resolve it to a single variable.

Also, $x$ & $y$ are positive integers.

$f\:=\:\frac{x}{y}+\frac{y}{x}+\frac{1}{xy}$

Edit : This one is simpler than the linked question and requires less steps.

Answer 1

$f, x, y$ are integers. (The proposition is not true if we allow $x,y$ to be rationals. e.g. let $x = y = \frac 13$ then $f = 11$)

$f = \frac {y}{x} + \frac {x}{y} + \frac {1}{xy} \implies xyf = x^2 + y^2 + 1$

$xyf \equiv x^2 + y^2 + 1 \pmod{3}$

If $x$ or $y$ are divisible by 3 then $xyf \equiv 0 \pmod 3.$

But then, it would not be possible for $x^2 + y^2 + 1 \equiv 0 \pmod 3$

If neither $x$ nor $y$ is $0$ modulo $3$ then $x^2 + y^2 + 1 \equiv 0 \pmod 3$

In which case $f\equiv 0 \pmod 3$

A quick primer on the field of integers mod $p$

We can think of modulo $p$ as the remainder we get when we divide by $p.$

If a number is equivalent to 0 mod p then it is divisible by $p.$

Modular arithmetic preserves addition and multiplication. That is, if we add to numbers and evaluate modulo p, we can just as easily evaluate the numbers modulo p before we add them. And, if we multiply integers and evaluate mod p, we can first evaluate modulo p and then multiply. If we get a result bigger than p we can evaluate that result modulo p.

We can use this to dramatically simplify problems, as there are only p cases to check for each variable in the equations, or for this problem 3 cases.

When we look at $x^2 + y^2 + 1$ we only have to consider: $1^2 \equiv 1 \pmod 3, 2^2 \equiv 1 \pmod 3$ and $0^2 \equiv 0 \pmod 3.$ If $x,y$ are not divisible by 3 then $x^2 + y^2 + 1 \equiv 0 \pmod 3$ If x or y are divisible by 3, then $x^2 + y^2 + 1 \equiv 1,2 \pmod 3$

If p is prime, the integers mod p form a ring. That is $ab\equiv 0\pmod p \implies$ p divides a or p divides b. Toward the end of the proof, we are using this property. We have shown that $xyf = x^2 + y^2 + 1 \equiv 0 \pmod 3$ and $x,y$ are not divisible by $3$ therefore $f$ is.

Answer 2

Hint $\bmod 3\!:\,\ {-}1 \equiv x^2-fxy+y^2\equiv (x+fy)^2\ $ if $\, f\equiv \pm1,\,$ so $\,f\equiv 0$

Answer 3

Write it as $$ xyf = x^2 + y^2 + 1 $$ Compute both sides mod $3$, that is, for $x,y=0,1,2$. It turns out that there are two cases:

• The LHS is zero and the RHS is not zero. This cannot happen.

• The LHS is not zero and the RHS is zero. This implies that $f \equiv 0 \bmod 3$.

$$ \begin{array}{cccc} x & y & xy & x^2 + y^2 + 1 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 2 & 0 & 2 \\ 1 & 0 & 0 & 2 \\ 1 & 1 & 1 & 0 \\ 1 & 2 & 2 & 0 \\ 2 & 0 & 0 & 2 \\ 2 & 1 & 2 & 0 \\ 2 & 2 & 1 & 0 \\ \end{array} $$