Today a friend told me the equality: $2^3 + 1 = 3^2$, and I wondered if there exist more solutions to the general problem $$x^y + 1 = y^x$$ where $x$ and $y$ are integers. Some research led me to the result for $x^y = y^x$, which has no integer solutions except $x = 2$ and $y = 4$ (assumed $x\neq y$). Is this a related result, or do more integer solutions exist?
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$1^2+1=2^1$. Can't think of anything else. – Quang Hoang Nov 28 '14 at 13:58
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1$x=0$ works too. – Macavity Nov 28 '14 at 13:59
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Sure $x = y$ is a solution to $x^y = y^x$ for all $x = y$. – Travis Willse Nov 28 '14 at 14:00
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maybe that could help : $e^{y\ln(x)} + 1 = e^{x\ln(y)}$. – Hedwig Nov 28 '14 at 14:01
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Yes, it is very similar to $x^y = y^x$ for integers $x$ and $y$. – Dietrich Burde Nov 28 '14 at 14:09
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3Hint: Even for very close values of x and y, the difference between $x^y$ and $y^x$ grows very powerfully $($assuming, of course, $x,y>1)$. So, if any solutions exist, they must be very small. – Lucian Nov 28 '14 at 19:25
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1This is also known as Catalan's Conjecture. I'm sure you can find formal proofs of that conjecture online – Brenton Jul 31 '15 at 14:07
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An alternate proof can be found here: http://math.stackexchange.com/questions/1609053/solving-a-diophantine-equation-with-2-variables. Does this make this question a duplicate? – S.C.B. Feb 18 '16 at 13:08
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2@Brenton No, Catalan's conjecture is far, far stronger (it allows both exponents to be arbitrary $>1$). It's true that it does solve the OP's question, but it is certainly not equivalent. Looking for formal proofs of Catalan's conjecture online would be unhelpful as any proof found would be thousands of times more complex than is needed to establish this special case. – Erick Wong Feb 22 '16 at 14:37
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Consider function $f:(0,\infty)\rightarrow \mathbb{R}, f(x)=(1+\frac{n}{x})^x$ which is strictly increasing and $f(x)<e^n$.
I. $x<y, y=x+n, n>0$, integer
$$\begin{align} x^y+1=y^x \implies & x ^{x+n}+1=(x+n)^x \\ \implies & x^n+\frac{1}{x^x}=(1+\frac{n}{x})^x<e^n \\ \implies & x<e \\ \implies & x=1, x=2\end{align}$$
For $x=1 \implies y=2$.
For $x=2 \implies 2^y+1= y^2 \implies y=3$
II. $x>y, x=y+k, k>0$, integer, using the idea of the situation I obtain the equation has no solution other than $(1,0)$.
The same method can solve equations:
$x^y=y^x$
$x^y+y=y^x+x$
$x^y+x=y^x+y$
where $x>0, y>0$ integers.
achille hui
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medicu
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You do have $$y^x-x^y=1$$ This is Catalan's theorem that the only solution are $(x,y)=(2,3),(1,2)$.
NOTE.- Catalan's Conjecture has becomed a theorem because Mihailescu has proved it recently enough.
Piquito
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