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

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    $\begingroup$ It is not a good practice to add a new part in your question after getting 2 answers. Ask it as a separate question. $\endgroup$ – SchrodingersCat Dec 25 '15 at 6:45
  • $\begingroup$ ok,I will add a new post. $\endgroup$ – user237685 Dec 25 '15 at 6:46
  • $\begingroup$ I'd rather add a link to that post: math.stackexchange.com/questions/1588433/… $\endgroup$ – Ivan Neretin Dec 25 '15 at 14:36

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

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.

  • $\begingroup$ It's Nice answer!+1 $\endgroup$ – user237685 Dec 25 '15 at 6:46
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    $\begingroup$ Good point, although I find your use of maximally divided confusing. You could just say that in the prime factorizations of both numerator and denominator $2$ must appear with even exponent… $\endgroup$ – A.P. Dec 25 '15 at 11:10

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.)

  • $\begingroup$ Are you forgetting about the divisors x,y,z, which may also be a multiple of 7? $\endgroup$ – ThomasMcLeod Dec 25 '15 at 6:59
  • $\begingroup$ No, and in fact I did mention them explicitly in the answer. They can decrease that $\alpha$ in $(supposedly\;2016)=\dots\cdot7^\alpha$, but still it remains a multiple of 7. $\endgroup$ – Ivan Neretin Dec 25 '15 at 8:02
  • $\begingroup$ Yeah now I get it. $\endgroup$ – ThomasMcLeod Dec 25 '15 at 22:27

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