Solution of $X^5=5 Y (Y+1)+1$ in integers.

Is it true that the

$$X^5=5 Y (Y+1)+1$$

has no non trivial solution (except the $X=1,Y=-1$ and $X=1,Y=0$) in integers. I tried to solve it as a quadratic equation on $Y$. I got that the $$\sqrt{5} \sqrt{4 X^5+1}$$ should be integer. But no idea how to continue from here.

Then I tried with Fermat Theorem $X^5\equiv X \pmod 5$ with hope to get some contradiction but again nothing useful.

EDITED: Just an observation from the @stuart stevenson answer: $X=10k+1$

So we have the following solutions:

$$Y\to \frac{1}{2} \left(\pm\sqrt{80000 k^5+40000 k^4+8000 k^3+800 k^2+40 k+1}-1\right)$$

Or the $$80000 k^5+40000 k^4+8000 k^3+800 k^2+40 k+1 = L^2 = \\ 1 + 40 k (1 + 20 k (1 + 10 k (1 + 5 k (1 + 2 k))))$$

This can be generalized as a recurrent product:

$$P_0(k)=2k+1 \\ P_n(k)=1+2^{n-1} 5kP_{n-1}(k)$$

And we assume that the $$P_4(k)\ne L^2$$ for any $k$.

EDITED:

I would assume that the $$X^p=p Y (Y+1)+1$$

has no non-trivial solution for any prime $p>2$.

EDITED:

Just some observation. It seems there are few $k$ for which the $$X^p=k Y (Y+1)+1$$

has solutions. For example for the $p=5$ and $p=7$ the $k$-s for which solution exists are $-1$ and $4$. And seems it has no solution for other $k$-s in the range $|k|<100$.

For the $p=3$ the $k$-s for which there is solutions are $\{-27, -21, -14, -7, -3, -1, 0, 1, 4, 6, 13, 27, 31\}$ for the range $|k|<50$ and seems for other $k$-s in this range there is no solution

Solution 1:

0. Overview
Following this old post, we attempt to solve the problem in the Number Field $K= \mathbb Q(\sqrt 5)$, which reduces to solving a set of $5$ Thue equations. We then solve them via PARI/GP which gives us the full solutions $(x,y)\in \{(1,-1),(1,0)\}$. In the following, let $w=\sqrt 5$ and $\alpha = (w+1)/2$.

1. Properties of $K$
$K$ has class number $1$, hence it is a PID and has unique factorization. As $5\equiv 1 \pmod 4$, the ring of integers is $R = \mathbb Z[(w+1)/2] = \mathbb Z[\alpha]$. The fundamental unit is $\eta = (w+ 1)/2 = \alpha$ and roots of unity is $\pm 1$. We remark that $\eta^{-1} = (w-1)/2$, as $$ \eta\eta^{-1} = \frac{w+1}{2}\frac{w-1}{2}=\frac{w^2-1}{4}=1 $$

2. Solving the original problem in $R$
We start by factorizing the original problem in $R$: $$ \begin{align*} (wy+\frac{w+1}{2})(wy + \frac{w-1}{2}) &= w^2y^2 + w(\frac{w+1}{2}+\frac{w-1}{2})y+\frac{w+1}{2}\frac{w-1}{2} \\ &= w^2 y + w^2 y + \frac{w^2-1}{4}\\ &= 5y^2 + 5y + 1 \\ &= x^5 \end{align*} $$ Denote $A = wy + (w+1)/2$ and $B = wy + (w-1)/2$, noting that $A,B\in R$. Since we have $$ A - B = 1, $$ $(A)+(B)$ generates $R$ and we conclude that $A,B$ are coprime.

Let the unique factorization of $x$ into irreducibles be $$ x = \pm \eta^k I_1^{e_1} I_2^{e_2} \cdots I_m^{e_m} $$ (where $I_j$ are the irreducibles). Then $$ AB = x^5 = \pm \eta^{5k} I_1^{5e_1} I_2^{5e_2} \cdots I_m^{5e_m} $$ Since $A,B$ are coprime, each irreducible power divides exactly one of $A$ or $B$. By relabelling $\{I_j\}$, we may assume that exactly the first $s$ irreducibles divide $A$, therefore $A$ has the form $$ A = \pm \eta^l I_1^{5e_1}I_2^{5e_2} \cdots I_s^{5e_s} = \pm \eta^l (a+b\alpha)^5,\;\;\;\;l,a,b\in\mathbb Z $$ since product of irreducibles are in $R$ and hence of the form $a+b\alpha$. We can factor $-1$ and powers of $\eta^5$ into $(a+b\alpha)^5$, hence a simplified form would be $$ A = \eta^l (a+b\alpha)^5,\;\;\; l,a,b\in\mathbb Z,\;\;\;\; 0\leq l\leq 4 $$ Remark: we can factor $\eta^{-5}$ into $(a+b\alpha)^5$ too since $\eta^{-1} = (w-1)/2$, which is why we can assume $0\leq l \leq 4$.

3. Solution via Thue equations
For our original problem to have a solution, there must be $y,l,a,b\in \mathbb Z$ (and $0\leq l\leq 4$) such that the equation below holds: $$ \begin{align*} wy + \frac{w+1}{2} &= A = \eta^l (a+b\alpha)^5 \\ \frac{1}{2} + (y+\frac{1}{2})w &= \left(\frac{w+1}{2}\right)^l \left(a + b\frac{w+1}{2}\right)^5 \\ \end{align*} $$

For each $l$ in $[0,4]$, equating the rational constant gives us a degree $5$ Thue equation, which we can solve using computational tools (using PARI/GP in our case). For example, letting $l=0$ we obtain: $$ \begin{align*} \frac{1}{2} + (y+\frac{1}{2})w &= \left( a^5 + \frac{5 a^4 b}{2} + 15 a^3 b^2+20 a^2 b^3 + \frac{35 a b^4}{2} + \frac{11 b^5}{2} \right) \\ &+ w\left( \frac{5 a^4 b}{2} + 5 a^3 b^2+10 a^2 b^3+ \frac{15 a b^4}{2}+ \frac{5 b^5}{2} \right) \\ \implies 1 &= 2 a^5+5 a^4 b+30 a^3 b^2+40 a^2 b^3+35 a b^4+11 b^5\\ \end{align*} $$ We solve this on the online PARI/GP calculator, via the command:

thue(thueinit(2*a^5 + 5*a^4 + 30*a^3 + 40*a^2 + 35*a + 11,1),1)

(The first '1' means to assume nothing.) This returns
%1 = []
which tells us that there are no solutions.

For the 4 remaining equations corresponding to $1\leq l\leq 4$, they are: $$ \begin{align*} l &= 1:& 1 &= a^5+15 a^4 b+40 a^3 b^2+70 a^2 b^3+55 a b^4+18 b^5 \\ l &= 2:& 1 &= 3 a^5+20 a^4 b+70 a^3 b^2+110 a^2 b^3+90 a b^4+29 b^5 \\ l &= 3: & 1 &= 4 a^5+35 a^4 b+110 a^3 b^2+180 a^2 b^3+145 a b^4+47 b^5 \\ l &= 4: & 1 &= 7 a^5+55 a^4 b+180 a^3 b^2+290 a^2 b^3+235 a b^4+76 b^5 \end{align*} $$ For convenience, the corresponding PARI/GP commands are
thue(thueinit(a^5+15*a^4+40*a^3+70*a^2+55*a+18,1),1)
thue(thueinit(3*a^5+20*a^4+70*a^3+110*a^2+90*a+29,1),1)
thue(thueinit(4*a^5+35*a^4+110*a^3+180*a^2+145*a+47,1),1)
thue(thueinit(7*a^5+55*a^4+180*a^3+290*a^2+235*a+76,1),1)

PARI/GP tells us that the only solutions are $(a,b) = (1,0)$ for $l=1$ and $(a,b) = (1,-1)$ for $l=4$. Putting in these values, we get the values of $A$ as $$ \begin{align*} (l,a,b) &= (1,1,0) &\implies A &= \left(\frac{w+1}{2}\right)^1(1)^5 = \frac{1}{2} + \frac{w}{2} \\ (l,a,b) &= (4,1,-1) &\implies A &= \left(\frac{w+1}{2}\right)^4 \left(1 - \frac{w+1}{2}\right)^5 = \frac{1}{2} - \frac{w}{2} \end{align*} $$ Recalling that $A=(1/2)+(y+1/2)w$, this in turn gives $y= 0$ and $y=-1$ respectively. Putting $y =0$ and $y=-1$ into our original equation, we get $x^5=1$ for both and hence $x=1$. This concludes that the only solutions are $$ (x,y) \in \{(1,-1),(1,0)\} $$

Remark: Perhaps it may be possible to solve the Thue equations manually so that this proof does not rely on computational tools.

Solution 2:

So $x^5 = 5y(y+1)+1$. We know then that $x^5 = 10 \frac{y(y+1)}{2}+1$ and the fractional part is integer as I have already said.

Therefore the units part of $x^5$ is $1$. We know (it's trivial to prove or you can look it up) that $x^5 \equiv x$ mod $10$. Therefore $x$ has unit part $1$ aswell.

Therefore $x = 10k+1$ where $k$ is an integer greater than $0$.

Not intended as a full answer but to demonstrate my commented ideas.