# $5^x - y^2 = 4$ Diophantine Equation

I wrote a Diophantine equation and tried solving it. Then I got stuck at a stage of the solution.

Problem: Find all $$(x,y)$$ positive integer pairs that satisfy the equation $$5^x - y^2=4$$.

My Partial Solution: If $$x$$ is an even number then $$x=2n\quad$$ ($$n \geq 1$$ an integer). Therefore $$(5^n - y)(5^n + y) = 4$$ and we can write $$\begin{equation*} \left\{ \begin{split} 5^x-y & = 1 \\ 5^x + y & = 4 \end{split} \right. \end{equation*}$$ Thus, $$2\cdot 5^x = 5$$ and there is no positive integer solution in this case.

If $$x$$ is an odd number,

$$\bullet$$ For $$x=1$$; $$\quad 5^1 -y^2=4 \implies y=1$$.

$$\bullet$$ For $$x=3$$; $$\quad 5^3 -y^2=4 \implies y=11$$.

$$\bullet$$ For $$x\geq 5$$; I thought of finding a contradiction using modular arithmetic. For example; $$x=2k + 3 , \quad$$ ($$k\geq 1$$ an integer) $$125\cdot 25^k - y^2 = 4$$. In $$\mod 24$$, $$5 - y^2 \equiv 4 \pmod{24}$$

But this is not a contradiction. So, I failed. How can I tell if the equation has a solution for $$x>3$$ or not? Thanks for your interest.

• Yeah, we we can say we multiply roughly by 25 giving $25\cdot 5^x-25\cdot y^2=100$ ... Commented Jun 27, 2021 at 15:43
• If $x=2n+1$ then we can write the above equation as $5(5^n-1)(5^n+1)=(y-1)(y+1)$ Commented Jun 27, 2021 at 15:58
• Recasting the equation as $5^x=y^2+4$, it is obvious that $y$ can only end in the digits $1,9$ or in other words $y \equiv \pm 1 \bmod 10$. Hence $y^2=100k^2 \pm 20k +1$ Commented Jun 27, 2021 at 16:31

## 2 Answers

$$5^{x} - y^2 = 4\tag{1}$$

We take the three cases $$x=3a, x=3a+1$$, and $$x=3a+2.$$
The problem can be reduced to finding the integer points on elliptic curves as follows.

$$\bullet\ x=3a$$
Let $$X=5^{a}$$, then we get $$y^2 =X^3 - 4.$$
According to LMFDB, this elliptic curve has integral solutions $$(X,y)=(2,\pm 2), (5,\pm 11).$$
From $$(5,\pm 11),$$ we get $$(x,y)=(3,11).$$

$$\bullet\ x=3a+1$$
Let $$X=5\cdot5^{a}, Y=5y$$, then we get $$Y^2 =X^3 - 100.$$
This elliptic curve has integral solutions $$(X,Y)=(5,\pm 5),(10,\pm 30),(34,\pm 198).$$
From $$(5,\pm 5)$$, we get $$(x,y)=(1,1).$$

$$\bullet\ x=3a+2$$
Let $$X=25\cdot5^{a}, Y=25y$$, then we get $$Y^2 =X^3 - 2500.$$
This elliptic curve has integral solution $$(X,Y)=(50,\pm 350).$$
We get no solution $$(x,y).$$

Hence there are only integral solutions $$(x,y)=(1,1),(3,11).$$

The number $$5^x$$ is either a perfect square or $$5$$ times a perfect square for any integer $$x$$.

Hence either $$m^2-y^2=4\ (1)$$ for $$m=5^k$$ or $$5n^2-y^2=4\ (2)$$ for $$n=5^l$$.

In the first equation, $$x=2k$$. The first equation is can be factored as $$(m-y)(m+y)=4=2^2,$$ so $$m$$ and $$y$$ must have the same parity and hence either $$m+y=2$$ and $$m-y=2$$ or $$m+y=-2$$ and $$m-y=-2$$. This gives $$(m,y)=(\pm2,0)$$, so $$\pm2=5^k$$, a contradiction.

In the second equation, $$x=2l+1$$. The second equation has integer solutions iff $$n$$ is the odd-indexed Fibonacci number. The equation $$2$$ is equivalent to $$5(5^l)^2-y^2=4.$$ Since the only Fibonacci numbers which are the powers of $$5$$ are $$1$$ and $$5$$, $$l$$ is either $$0$$ or $$1$$ and hence $$(x,y)\in\{(1,1), (3,11)\}$$ since $$y$$ is positive.