Solving a system of two equations

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.