# Find a example such $\frac{(x+y)^{x+y}(y+z)^{y+z}(x+z)^{x+z}}{x^{2x}y^{2y}z^{2z}}=2016$

Assume $x,y,z$ be postive integers,and Find one example $(x,y,z)$ such
$$\dfrac{(x+y)^{x+y}(y+z)^{y+z}(x+z)^{x+z}}{x^{2x}y^{2y}z^{2z}}=2016$$

#### Solutions Collecting From Web of "Find a example such $\frac{(x+y)^{x+y}(y+z)^{y+z}(x+z)^{x+z}}{x^{2x}y^{2y}z^{2z}}=2016$"

$2016 = 2^5*3^2*7$

So now we wish to find

$$x,y,z \in \Bbb{N}$$

such that

$$\frac{(x+y)^{x+y}(y+z)^{y+z}(x+z)^{x+z}}{x^{2x}y^{2y}z^{2z}} = 2^5*3^2*7$$

Clearly at least one of $(x+y), (y+z)…$ is going to be even (to get that power of 2), But look carefully at the power of 2, it has an odd exponent

Observe that the numerator can be a product of the form

$$\text{odd}^{\text{odd}} \times \text{odd}^{\text{odd}} \times \text{odd}^{\text{odd}}$$
$$\text{even}^{\text{even}} \times \text{odd}^{\text{odd}} \times \text{odd}^{\text{odd}}$$
$$\text{even}^{\text{even}} \times \text{even}^{\text{even}} \times \text{even}^{\text{even}}$$

(up to permutation), and the denominators necessarily must be of the form

$$\text{odd}^{\text{even}} \times \text{odd}^{\text{even}} \times \text{odd}^{\text{even}}$$
$$\text{even}^{\text{even}} \times \text{odd}^{\text{even}} \times \text{odd}^{\text{even}}$$
$$\text{even}^{\text{even}} \times \text{even}^{\text{even}} \times \text{even}^{\text{even}}$$

So if the numerator is to be maximally divided by $2^k$ term this term will have even k.

And if the denominator is to be maximally divided by $2^j$ this term will also have an even $j$.

So this expression, is only divisible by $2^{j-k}$ where $j-k$ is even, yet here we claim it results in $2^5$ which has an odd exponent, a contradiction.

Thus we conclude there is no solution.

Such a thing will never happen. Why? Because 2016 is divisible by 7. Let’s see what can be said about the power of 7 in the prime decomposition of this expression. Those of $x,\;y,\;z,\;x+y,\;y+z,\;z+x$ which are not divisible by 7 themselves, contribute nothing. Those which are, contribute (add or detract) a multiple of themselves, and hence a multiple of 7. But $2016=2^5\cdot3^2\cdot7^1$, and 1 is not a multiple of 7.

(The same reasoning could be applied to 2 or 3, of course.)