10
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.


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.


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.

$\endgroup$

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.