14
$\begingroup$

Prompted by this question, I tried to show that $(2,5)$ is the only solution in positive integers of $x^y - y^x = x+y$ (which would show, a fortiori, that it's the only solution in primes). It's convenient to rewrite the equation as $f(x,y) = x^y - y^x - x - y = 0$.

With the aid of some trial calculations, I reasoned informally as follows:

If $x=1$ then $f(x,y) = -2y=0$, implying $y=0$.

If $x=2$ and $y \leq 5$, then a case-by-case check shows that only $(2,5)$ is a solution.

If $(x,y) = (2,6)$ then $f(x,y) = 20$, and as $y$ increases above $6$, $f(x,y)$ increases.

If $x \geq 3$ and $y \leq x$ then $f(x,y) < 0$.

If $x \geq 3$ and $y=x+1$, then $f(x,y)>0$, and as $y$ increases above $x+1$, $f(x,y)$ increases.

How can the above be made into a rigorous proof? I've included calculus as a tag since it could be useful in showing under what conditions $f(x,y)$ is an increasing function of $y$ (viewing it as a real variable).

$\endgroup$
11
$\begingroup$

If $(x,y) = (2,6)$ then $f(x,y) = 20$, and as $y$ increases above $6$, $f(x,y)$ increases.

To prove this rigorously, let us prove by induction that $$2^{X-1}\gt X+1\tag1$$ for $X\ge 6$.

The base case : $2^{6-1}=32\gt 7=6+1$.

Supposing that $(1)$ holds for some $X$ gives $$2^{(X+1)-1}=2\cdot 2^{X-1}\gt 2(X+1)=X+X+2\gt (X+1)+1\quad\blacksquare$$

Using $(1)$, we get, for $y\ge 6$, $$f(2,y+1)-f(2,y)=2(2^{y-1}-y-1)\gt 0$$

If $x \geq 3$ and $y \leq x$ then $f(x,y) < 0$.

  • $f(x,1)=-2\lt 0$.

  • To prove that $f(x,2)\lt 0$ for $x\ge 3$, let us prove by induction that $$2^X\gt X^2-X-2\tag2$$ for $X\ge 4$. The base case : $2^4=16\gt 10=4^2-4-2$. Supposing that $(2)$ holds for some $X$ gives $$\begin{align}2^{X+1}&=2\cdot 2^X\\&\gt 2(X^2-X-2)\\&=(X+1)^2-(X+1)-2+X(X-3)-2\\&\ge (X+1)^2-(X+1)-2+4(4-3)-2\\&\gt (X+1)^2-(X+1)-2\quad\blacksquare\end{align}$$ Also, it is easy to see that $(2)$ holds for $X=3$. Using this, we get, for $x\ge 3$, $$f(x,2)=(x^2-x-2)-2^x\lt 0$$

  • To prove that $f(x,y)\lt 0$ for $3\le y\le x$, we use the fact that $y=\frac{\ln x}{x}$ is decreasing for $x\gt e$. We see that $\frac{\ln y}{y}\ge \frac{\ln x}{x}$ for $3\le y\le x$ from which $x^y-y^x\le 0$ follows. It follows from this that we have, for $3\le y\le x$, $$f(x,y)=x^y-y^x-x-y\le 0-x-y\lt 0$$

If $x \geq 3$ and $y=x+1$, then $f(x,y)>0$, and as $y$ increases above $x+1$, $f(x,y)$ increases.

This answer proves that $f(x,y)\gt 0$ for $x\ge 3$ and $y\ge x+2$.

Let $y=x+a$ where $a\ge 2$ is an integer. Then,

$$\frac{f(x,x+a)}{x^x}=x^{a}-\left(1+\frac ax\right)^x-\frac{2}{x^{x-1}}-\frac{a}{x^x}\gt x^a-e^a-\frac 29-\frac{a}{x^x}:=g(x)$$

Since $g'(x)=ax^{a-1}+\frac{a+a\ln x}{x^x}\gt 0$, $g(x)$ is increasing with $g(3)=3^a-e^a-\frac 29-\frac{a}{27}:=h(a)$.

We have $h'(a)=3^a\ln 3-e^a-\frac{1}{27}$ and $h''(a)=3^a(\ln 3)^2-e^a\gt 0$ with $h'(2)=9\ln 3-e^2-\frac{1}{27}\gt 0$ and $h(2)=9-e^2-\frac 29-\frac{2}{27}\gt 0$.

So, $h(a)\gt 0$ for $a\ge 2$. It follows from this that $g(x)\gt 0$ for $x\ge 3$.

Therefore, $f(x,x+a)\gt 0$ follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.