Differentiation in integer solutions

Question

What would you suggest as examples to demonstrate as applications of differentiation in finding integer solutions of an equation for advanced level students?
Here you have one example which I have come across so far relevant to the mentioned issue.
To prove that there is only one positive integer pair $( n,m )$ satisfying $n^m = m^n$ we can use the knowledge of stationary points of the graph $y = (ln x)/x$ . Since the only stationary point is at $x = e (>2)$ and it is a maximum we can say that one value of the pair $(n,m)$ is $2$. Hence we can find the other value of the pair.
Edit ; $n , m $ are distinct values
Here is the generalized version which I could able to construct,
How can you prove that there is only one triplet $(n,m,p)$ of positive integers satisfying the equation $$n^{m^p} = m^{n^p}$$ given that $ m ≠ n $. I would like to know different types of examples.

Answer 1

I think the scope of your question is too narrow. If you ask more broadly about the use of calculus in number theory (not just differentiation for solving Diophantine equations) then books on analytic number theory will provide lots of examples. For instance, there is an easy upper bound $\log(n!) < n\log n$ for $n \geq 2$, and with calculus we can derive a related lower bound $\log(n!) > n\log n - n$: see Theorem 2.1 here.

In a different direction, if we want to solve a Diophantine equations in polynomials rather than in integers, then we can use differentiation to prove things that are far more difficult in the setting of integers. Consider the polynomial analogue of Fermat's Last Theorem: for $n \geq 3$, the only polynomial solutions to the equation $f(x)^n + g(x)^n = h(x)^n$ for $f(x), g(x)$, and $h(x)$ in $\mathbf R[x]$ or $\mathbf C[x]$ are constant solutions. This can be proved by differentiating both sides of the Fermat equation. See the first two pages of a paper by Granville and Tucker here.

Answer 2

Here is a similar but slightly different example:

Show that there are no distinct integers $m$ and $n$ satisfying $e^n-n=e^m-m$.

Solution: Suppose there are such integral solutions $m<n$. Then applying the Mean Value Theorem to the function $e^x$ on the interval $[m,n]$, we see that there exists a number $c$ between $m$ and $n$ such that $e^c=1$. But then we must have $c=0$. This shows that $m<0<n$. Now, from $e^n-n=e^m-m$ it follows $e^n-e^m=n-m$. Multiplying both sides by $e^{-m}$ we obtain: $$e^{n-m}-1=(n-m)e^{-m}\ \ \mathrm{or}\ \ e^{n-m}-(n-m)e^{-m}-1=0.$$ In other words, $e$ satisfies the algebraic equation $x^{n-m}-(n-m)x^{-m}-1=0$, which is a contradiction, as $e$ is not algebraic. Note that since $m<0$, the exponent of $x$ in $x^{-m}$ is positive.

More generally: The equation $p^n-n=p^m-m$ does not have any integral solutions in $p,m,n$ satisfying $p>2$ and $m\neq n$. (For $p=2$ there is a solution $m=0, n=1$.)

The original equation $e^n-n=e^m-m$ could be solved without using any calculus, as pointed out by fedja in a comment, only using the fact that $e$ is not algebraic. To show that the equation $p^n-n=p^m-m$ does not have any integral solutions in $p,m,n$ satisfying $p>2$ and $m\neq n$, suppose there are such integral solutions $p>2$ and $m<n$. As $p>2$, it is easy to see that $m\neq0$. Then applying the Mean Value Theorem to the function $p^x$ on the interval $[m,n]$, we see that there exists a number $c$ between $m$ and $n$ such that $p^c=1$. But then we must have $c=0$. This shows that $m<0<n$. Now, from $p^n-n=p^m-m$ it follows $p^n-p^m=n-m$. Multiplying both sides by $p^{-m}$ we obtain: $$p^{n-m}-1=(n-m)p^{-m}\ \ \mathrm{or}\ \ p^{n-m}-(n-m)p^{-m}=1.$$ This means $p^{n-m}$ and $p^{-m}$ are coprime, which is absurd, as $n-m>0$ and $-m>0$.