Solve in integers $(x^2+2)(y^2+3)(z^2+4)=60xyz$

Question

Solve in integers the equation $$(x^2+2)(y^2+3)(z^2+4)=60xyz$$

My try:

We have $x,y,z \ne 0$ So we can write the equation as:

$$\left(x+\frac{2}{x}\right)\left(y+\frac{3}{y}\right)\left(z+\frac{4}{z}\right)=60$$

Assuming $x,y,z$ are positive integers

By AM-GM

we have

$$x+\frac{2}{x} \gt 2\sqrt{2}$$

$$y+\frac{3}{y}\gt 2\sqrt{3}$$

$$z+\frac{4}{z}\ge 4$$

Any idea from here?

Answer 1

I suppose you need a slightly modified version of the inequalities. Recall that all $x,y,z$ are integers, and in fact, we can think of them as of naturals.

Then, for any $x,y,z \in \mathbb{N}$ and $y \neq 2$, $z \neq 2$, $z\neq 3$, we have $$ x+\frac{2}{x}\geq 3,\quad y+\frac{3}{y} \geq 4, \quad z+\frac{4}{z} \geq 5. $$ Therefore, if $y\neq 2$, $z \neq 2$ and $z \neq 3$, we require all inequalities become equalities since $60 = 3\cdot 4 \cdot 5$, and so solutions are $$ (x,y,z) \in \{1,2\} \times \{1,3\} \times \{1, 4\} $$ modulo all the required sign changes (since $x,y,z$ might be negative integers as well).

The remaining cases are when $y=2$, $z=2$ or $z=3$. It shouldn't be hard, and I guess, these imply some contradictions, need to check it, which I never did!

Answer 2

It is obvious without even doing any arithmetic that if $z=1000$, there is nothing that $x$ or $y$ could possibly be to overcome the huge number that is $z^2$.

The LHS grows in $O(n^2)$ and the RHS grows only in $O(n)$. That tells you that there is a computable upper limit to how large the (positive) values of $x,y,z$ can be. The question can be reduced to asking "when does the faster growth of $O(n^2)$ overtake the larger overhead factor of $60$?"

If $x=y=1$ then the largest $z$ can be is $4$.

If $y=z=1$ then the largest $x$ can be is $2$.

If $z=x=1$ then the largest $y$ can be is $3$.

Now the problem can be solved with exhaustion.

Answer 3

Consider mod 5: $$x\equiv 0,1,2,3,4\pmod{5} \Rightarrow x^2+2\equiv 2,3,1,1,3\pmod{5}\\ y\equiv 0,1,2,3,4\pmod{5} \Rightarrow y^2+3\equiv 3,4,2,2,4\pmod{5}\\ z\equiv 0,1,2,3,4\pmod{5} \Rightarrow z^2+4\equiv 4,\color{red}0,3,3,\color{red}0\pmod{5}$$

So, $z$ can be $\pm 1,\pm 4$ and cannot be $\pm 6$ or higher, because for $z=\pm 6$: $$(x^2+2)(y^2+3)>2\sqrt{2}\cdot 2\sqrt{3}xy\approx 9.8xy>9xy.$$ For $z=1,4$: $$(x^2+2)(y^2+3)=12xy \Rightarrow (x,y)=\pm(1,1),\pm(1,3),\pm(2,1),\pm (2,3).$$ For $z=-1,-4$: $$(x^2+2)(y^2+3)=-12xy \Rightarrow (x,y)=\pm(-1,1),\pm(-1,3),\pm(-2,1),\pm(-2,3).$$ Note: For the equation of $x$ and $y$, you can consider mod 4: $$x\equiv 0,1,2,3\pmod{4} \Rightarrow x^2+2\equiv 2,3,2,3\pmod{4}\\ y\equiv 0,1,2,3\pmod{4} \Rightarrow y^2+3\equiv 3,\color{red}0,3,\color{red}0\pmod{4}$$ Can you collect the $32$ solutions?

Refer to WA answer.