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I have a system of equations: $$ \begin{cases} x\cdot y=6 \\ x^y+y^x=17 \end{cases} $$ I was able to guess that the pair $2,3$ satisfies the system, but my question is: how to solve such system of equations OR how to prove that this pair is the only solution?

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  • $\begingroup$ I've done it. ;D $\endgroup$ – VanDerWarden Aug 25 '15 at 19:25
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Given that the tag is Diophantine-equations we can only have the pairs

$$ (1,6), (2,3), (3,2), (6,1), \tag 1 $$

as

$$ x \cdot y = 6. \tag 2 $$

Put it in and we get

$$ \begin{eqnarray} (1,6) - (6,1) &:& x^y + y^x = 1 + 6 = 7\\\\ (2,3) - (3,2) &:& x^y + y^x = 8 + 9 = 17 \end{eqnarray} \tag 3 $$

So the solutions

$$ \bbox[16px,border:2px solid #800000] { (2,3) - (3,2) } \tag 4 $$

are the only ones.

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  • $\begingroup$ from where do you know that this is a Diophantine equation? $\endgroup$ – Dr. Sonnhard Graubner Aug 25 '15 at 19:04
  • $\begingroup$ It is tagged as Diophantine equation. $\endgroup$ – johannesvalks Aug 25 '15 at 19:05
  • $\begingroup$ Sorry, my mistake. It's not a Diophantine equations, but thanks anyway for your clever contribution. $\endgroup$ – VanDerWarden Aug 25 '15 at 19:06
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HINT: from the first equation we get $$y=\frac{6}{x}$$ plugging this in the second one we get $$x^{6/x}+\left(\frac{6}{x}\right)^x=17$$ this equation can be solved by a numeric method

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  • $\begingroup$ this does not prove that the solution is unique $\endgroup$ – Surb Aug 25 '15 at 18:42
  • $\begingroup$ yes this is right this equation has two solutions, we must do more to prove this $\endgroup$ – Dr. Sonnhard Graubner Aug 25 '15 at 18:48
  • $\begingroup$ But how to solve the second equation you wrote? Any hints? $\endgroup$ – VanDerWarden Aug 25 '15 at 18:59
  • $\begingroup$ use a numerical method e.g. the Newton method $\endgroup$ – Dr. Sonnhard Graubner Aug 25 '15 at 19:00
  • $\begingroup$ and plot the graph of the equation to find start values for the Newton method $\endgroup$ – Dr. Sonnhard Graubner Aug 25 '15 at 19:01
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I have an idea for uniqueness. Suppose that we have a solution (a,b). Then by second equation (b,a) must be also a solution. Then by equation one we have
$$ab = ba = 6$$ Then (a,b) and (b,a) are symmetric. Which means if there is a solution, there must be at least one more solution which is symmetric to other. Namely if (2,3) is solution, then (3,2) is also solution. So solution is not unique.

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  • $\begingroup$ this makes sense, but I guess OP meant are there other solution than $(2,3)$ up to symmetry of the solution $\endgroup$ – Surb Aug 25 '15 at 18:55
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    $\begingroup$ If this is the case, OP needs to clarify the question. Lets wait for him to speak. $\endgroup$ – Salihcyilmaz Aug 25 '15 at 18:59

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