Exponential diophantine equation involving a prime number

Question

Few days ago I found a question which was unfortunately put "on hold" and later it was even deleted. The question asked to solve the following exponential diophantine equation $$p^{p+1}+(p+1)^p=x^2$$ where $x$ is a positive integer and $p$ is a prime number. Since I like diophantine equations I found this one very interesting and I tried to solve it.

Here is what I did: we can assume that $p$ is odd because for $p=2$ there is no solution, so $p+1=2n$ and thus we can write $$(2n)^p=(p+1)^p=x^2-p^{2n}=(x+p^n)(x-p^n).$$

Now, from the original equation we get that $x$ is odd and $x\equiv \pm 1\pmod p$, then $\gcd(x+p^n, x-p^n)=2$, and therefore we can write $x+p^n=2^{\alpha}\cdot y$ and $x-p^n=2^{\beta}\cdot z$, with exactly one of $\alpha$ and $\beta$ being equal to $1$ because $\alpha+\beta\ge p\ge 3$. More precisely, if $n=2^r\cdot m$, then $$2^{p+rp}\cdot m^p=2^{\alpha+\beta}\cdot yz.$$

Hence $p(r+1)=\alpha+\beta$ and $m^p=yz$ with $\gcd(y,z)=1$, which means that $y=y_1^p$ and $z=z_1^p$, for some $y_1, z_1\in\Bbb{Z}^+$ such that $\gcd(y_1, z_1)=1$.

Sadly, I couldn't do anything else and I got stuck here until I remembered that the general equation $y^x+x^y=z^2$ was solved in this paper when $\gcd(x,y)=1$, $xy$ is even and $\min\{x,y\}>1$.

In our particular equation the required conditions are satisfied because $\gcd(p+1,p)=1$, $(p+1)p$ is even and $\min\{p+1,p\}\ge 3$, so according to that paper our equation doesn't have any solutions. The only problem that I found is that the equation $y^x+x^y=z^2$ was solved using bounds for linear forms in logarithms, which are beyond from being elementary.

So this is my question: is there a more elementary solution for the equation $p^{p+1}+(p+1)^p=x^2$ ? Thanks in advance for your answers.

Answer 1

As you noted: $$2^pn^p=(x+p^n)(x-p^n)$$

$$2^{p-2}n^p=\frac{x+p^n}2\frac{x-p^n}2$$ With the factors on the RHS being coprime and one of them odd and the other even.

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If the first is the even one then $$\frac{x+p^n}2=2^{p-2}a^p$$ $$\frac{x-p^n}2=b^p$$ For integers $a,b$ with $ab=n$ because if a $k-$th power is the product of coprime factors, each of the factors is a $k-$th power. Then $$a^p=\frac{p^n+\sqrt{p^{p+1}+(p+1)^p}}{2^{p-1}}$$ $x\equiv\pm 1\pmod p$ so $a\equiv \pm 1$ and $$1<\left(\frac{p^n+\sqrt{p^{p+1}+(p+1)^p}}{2^{p-1}}\right)^{\frac 1p}=a<p-1.$$ For $p\geq 3$, and there's no number on this region that can satisfy $a\equiv \pm 1$. Contradiction.

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If the second factor is the even one the proof is almost exactly the same $$\frac{x+p^n}2=a^p$$ $$\frac{x-p^n}2=2^{p-2}b^p$$ $$1<\left(\frac{p^n+\sqrt{p^{p+1}+(p+1)^p}}2\right)^{\frac 1p}=a<p-1$$ $$a\equiv x\equiv \pm 1\pmod p.$$ Contradiction.