Find short and simple methods to solve $24x^4+1=y^2$

Question

Find all this diophantine equation $$24x^4+1=y^2\tag{1} $$ postive integers solution

it is clear $(x,y)=(1,5)$

I know $y^{2}=Dx^{4}+1$, where $D>0$ and is not a perfect square, has at most two solutions in positive integers (cf. L. J. Mordell, Diophantine equations, p. 270.

Does this equation have another proof such Lucas's assertion, with short and simple methods? Like this paper: Anglin, W. S. "The Square Pyramid Puzzle." Amer. Math. Monthly 97, 120-124, 1990. The square pyramid puzzle

In the paper,Following two question have simple methods to solve it.

There are no positive integers $x$ such $2x^4+1$ is a square.

and

There is exactly one positive integer $x$,namely $1$, such that $8x^4+1$ is a square?

But How can I find simple methods to solve $(1)$?

Answer 1

To begin with:

$$24x^4+1=y^2 \Leftrightarrow 6x^4=\frac{y-1}{2}\cdot\frac{y+1}{2}$$

let $\frac{y-1}{2}=a$. Then we have $6x^4=a(a+1)$; $gcd(a,a+1)=1$, so we have $2$ cases: $\{a,a+1\}=\{3u^4,2v^4\}$ or $\{a,a+1\}=\{u^4,6v^4\}$ (because $p^{4v_p(x)}$ divides $a$ or $a+1$)

Easy cases:

Case 1:

$\{a,a+1\}=\{u^4,6v^4\}$.

If $u^4-6v^4=1$. If $u$ is odd, then we get a contradiction $\pmod{8}$ (we get $1\equiv -5\pmod{8}$). So let $v=2k$. Then, $24k^4+1=u^4$. So we got another solution for our initial equation. From $(x,y))$ we got $(k,u^2)$ which is, in terms of $x$ and $y$, $\bigg(\big(\frac{y+1}{48}\big)^\frac{1}{4},\big(\frac{y-1}{2}\big)^\frac{1}{2}\bigg)$, so from infinite descent, we will keep going and reach 3 situations:

• we keep going through this case which will lead to a contradiction because we will get infinitely small solutions
• we end up in any other case, except the hard one(which will lead to a contradiction again because we have shown so using some modular arithmetic
• or we end up in the hard case, which will lead to a contradiction again, because if $\bigg(\big(\frac{y+1}{48}\big)^\frac{1}{4},\big(\frac{y-1}{2}\big)^\frac{1}{2}\bigg)$ leads to the hard case, then clearly for $\bigg(\big(\frac{y+1}{48}\big)^\frac{1}{4},\big(\frac{y-1}{2}\big)^\frac{1}{2}\bigg)\in\mathbb{N}\times\mathbb{N}$ we must have $y\equiv 5\pmod{6}$, but we also have $\frac{\sqrt{\frac{y-1}{2}}+1}{2}=3v^4$ so $\sqrt{\frac{y-1}{2}}+1\equiv 0\pmod{6}$ so $\sqrt{\frac{y-1}{2}}\equiv 5\pmod{6}$ so $\frac{y-1}{2}\equiv 1\pmod{6}$ so $y\equiv 3\pmod{6}$, contradiction

So we cannot have $u^4-6v^4=1$.

If $6v^4-u^4=1$, we get a contradiction$\pmod{3}$ (we get $1\equiv -1\pmod{3}$)

So we cannot have $6v^4-u^4=1$ either.

Case 2:

$\{a,a+1\}=\{3u^4,2v^4\}$.

If $2v^4-3u^4=1$, we get a contradiction$\pmod{3}$ (we get $1\equiv 2\pmod{3}$).

So we cannot have $2v^4-3u^4=1$

So we have $3u^4-2v^4=1$, the hard case. Before I actually begin discussing the hard case, I want to say that we will use brute force and a lot of calculations for this. Why? Well, the classical methods for solving diophantine equations are the following:

• modular arithmetic which, in our case, I think is useless, because $3-2\equiv 1\pmod{p}$ and we can control the primes that divide $u$ and $v$, so $-3$ and $2$ are quadratic residues etc. In fewer words, i really do not think these methdos work because how versatile this equation is while analyzing $\pmod{p}$
• maybe using the Pythagorean triplet form, but I do not know how to reach a Pythagorean equation from $3u^4-2v^4=1$
• Using some interesting substitutions and/or decompositions which again, I cannot seem to find

So because you want an elementary easy solution, I think that the only approach which will at least lead us in the direction of the solutions is the following:

• we calculate the solutions of $3v^2-2u^2=1$ and see why $u=v=1$ is the only solution for which $u$ and $v$ are perfect squares, giving us that the only solution to $3u^4-2v^4=1$ is $u=v=1$, which gives us $x=1$ and $y=5$

The hard case:

If $(u_0,b_0)$ is the fundamental solution (i.e. the smallest non trivial solution) of $ax^2-bx^2=1$, then the general solutions are:

$$x_n=x_0\frac{1}{2}\bigg[\big(u_0+v_0\sqrt{ab}\big)^n+\big(u_0-v_0\sqrt{ab}\big)^n\bigg]+by_0\frac{1}{2\sqrt{ab}}\bigg[\big(u_0+v_0\sqrt{ab}\big)^n-\big(u_0-v_0\sqrt{ab}\big)^n\bigg]$$

and $$y_n=y_0\frac{1}{2}\bigg[\big(u_0+v_0\sqrt{ab}\big)^n+\big(u_0-v_0\sqrt{ab}\big)^n\bigg]+ax_0\frac{1}{2\sqrt{ab}}\bigg[\big(u_0+v_0\sqrt{ab}\big)^n-\big(u_0-v_0\sqrt{ab}\big)^n\bigg]$$

Where $(u_0,v_0)$ is the fundamental solution of $u^2-abv^2=1$ and $(x_0,y_0)$ is the fundamental solution of $ax^2-ay^2=1$

The fundamental solution of $u^2-6v^2=1$ is $(5,2)$ and the fundamental solution of $3u^2-2v^2=1$ is $(1,1)$. So the general solutions for $3x^2-2y^2=1$ are

$$x_n=\frac{1}{2}\bigg[\big(5+2\sqrt{6}\big)^n+\big(5-2\sqrt{6}\big)^n\bigg]+\frac{1}{\sqrt{6}}\bigg[\big(5+2\sqrt{6}\big)^n-\big(5-2\sqrt{6}\big)^n\bigg]$$

and

$$y_n=\frac{1}{2}\bigg[\big(5+2\sqrt{6}\big)^n+\big(5-2\sqrt{6}\big)^n\bigg]+\frac{5}{2\sqrt{6}}\bigg[\big(5+2\sqrt{6}\big)^n-\big(5-2\sqrt{6}\big)^n\bigg]$$

Now, we can deduce some stuff in we analyze $\pmod{9}$ by expanding the above forms using the binomial theorem. I will not include the calculations here, because they are long. Anyhow, I checked with an engine, and we have the following:

$$n\equiv0\pmod{6}\Rightarrow x\equiv1\pmod{9},y\equiv1\pmod{9}$$ $$n\equiv1\pmod{6}\Rightarrow x\equiv0\pmod{9},y\equiv2\pmod{9}$$ $$n\equiv2\pmod{6}\Rightarrow x\equiv2\pmod{9},y\equiv1\pmod{9}$$ $$n\equiv3\pmod{6}\Rightarrow x\equiv2\pmod{9},y\equiv2\pmod{9}$$ $$n\equiv4\pmod{6}\Rightarrow x\equiv0\pmod{9},y\equiv7\pmod{9}$$ $$n\equiv5\pmod{6}\Rightarrow x\equiv1\pmod{9},y\equiv2\pmod{9}$$

So we can exclude $n\equiv 1,2,3,5\pmod{6}$, because in those cases, one of $x$ and $y$ is $2\pmod{9}$, so it cannot be a square.

However, I think this is the last of my efforts. Take, for example, $(x_{12},y_{12})$. It is $(804062262961,984771132841)$ Both these numbers are $\equiv 1\pmod{9},\pmod{5},\pmod{8},\pmod{11},\pmod{97}$. Sincerely, I do not know how to tackle those numbers, and the ones that may appear as $n\equiv 0,4\pmod{6}$. No modular reasoning works.

To conclude:

I hope I helped you, I tried every elementary approach I could, and found no apparent repeating contradiction even with some computational power. This $3v^4-2u^4=1$ seems.. very hard if not impossible to prove with elementary methods. I hope I did not oversee an easy approach (but i don't think so, as all other answers didn't show any progress for case $3v^4-2u^4=1$).

Farewell

Answer 2

$\frac{y-1}{2} \cdot \frac{y+1}{2} = 6x^4$

So $\frac{y+1}{2} = p a^4, \frac{y-1}{2} = q b^4$ where $pq = 6$ and we need to solve the equation $p a^4 - q b^4 = 1

Case 1: $p = 6, q = 1$

This is impossible modulo 3.

Case 2: $p = 2, q = 3$

This is impossible modulo 3.

Case 3: $p = 1, q = 6$

We will show that there is no solution using the method of infinite descent. Take the minimal solution in positive integers $a, b$.

Moving sides and factoring we get $\frac{a - 1}{2} \cdot \frac{a+1}{2} \cdot \frac{a^2 + 1}{2} = 12 (\frac{b}{2})^4$, therefore there exist coprime positive integers $\alpha, \beta, \gamma$ and coprime integers $m,n,k$ such that $\frac{a-1}{2} = \alpha m^4, \frac{a+1}{2} = \beta n^4, \frac{a^2 + 1}{2} = \gamma k^4$ such that $\alpha \beta \gamma = 12$.

Notice that $\gamma | \frac{a^2 + 1}{2}$, which means that neither $2$ nor $3$ can divide $\gamma$, so we must have $\gamma = 1$ and so $\alpha \beta = 12$. Therefore, $\frac{a^2 + 1}{2} - 2 \cdot \frac{a+1}{2} \cdot \frac{a-1}{2} = k^4 - 24(mn)^4 = 1$

Now we can repeat the same argument again: $\frac{k-1}{2} \cdot \frac{k+1}{2} \cdot \frac{k^2 + 1}{2} = 3(mn)^4$, so we get $\frac{k^2 + 1}{2} = u^4$, and $\frac{k - 1}{2}, \frac{k + 1}{2}$ are equal to $3v^4, w^4$ in some order. From this we get $u^4 - 6(vw)^4 = 1$, which is a smaller solution to our original equation, a contradiction.

Case 4: $p = 3, q = 2$

In this case we have to solve the equation $3a^4 - 2b^4 = 1$. Unfortunately, I do not know of a proof of this fact which is as simple as Case 3 and entirely elementary (nor am I sure such a proof exists). However there is this paper by R.T. Bumby, which solves the more general equation $3x^4 - 2y^2 = 1$ (which has besides the trivial solution $(1,1)$ also the more surprising $(3, 11)$) using essentially elementary methods, relying only on unique factorization in $\mathbb{Z}[\sqrt-2]$.

I have only skimmed this paper by Paolo Ribenboim, but it claims to give an algorithm to find all solutions to the equation $x^2 - Dy^4 = 1$ with fixed $D$ apparently with an elementary proof.

This article is a great survey by P.G. Walsh on questions like these about Pell equations: it contains good explanations about many results and probably contains all the references you'll run across studying this topic.

Hope this answer is of help.

Answer 3

Theorem. The number $24x^4+1$ is square only if $x=0,1$.

Proof. Suppose that $24x^4+1=y^2$, $y=2s+1$. Then $6x^4=s(s+1)$.

1) If $s=$even$=2s_1$, then $3x^4=s_1(2s_1+1)$.

i) If $3|s_1$, then $s_1=3s_2$, then $x^4=s_2(6s_2+1)$. But $(s_2,6s_2+1)=1$. Hence exists $u,v$ such that $x^4=u^4v^4$ and $s_2=u^4$, $6s_2+1=v^4$. Hence $6u^4+1=v^4$. But when $u\equiv 1,2,3,4(\textrm{mod}5)$ and $v\equiv 0,1,2,3,4(\textrm{mod}5)$, then $6u^4+1-v^4\neq0(\textrm{mod}5)$. Hence $u\equiv0(\textrm{mod}5)\Rightarrow v^4=6\cdot 5^4 k^4+1$ and $v$ is odd. Hence $v^2=2v_1+1\Rightarrow v^4=4v_1^2+4v_1+1=6u_1^4+1$, where $u_1=5k$. Hence $$ 2v_1(v_1+1)=3u_1^4.\tag 1 $$ Now if $v_1=$even, then $(2v_1,v_1+1)=1$. Also if $3|v_1$, then $v_1=6v_2$ and $12v_2(6v_2+1)=3u_1^4$. Hence $4v_2(6v_2+1)=u_1^4$, with $(4v_2,6v_2+1)=1$. Hence exists $u',v'$ such that $(u')^4=4v_2$ and $(v')^4=6v_2+1$. Hence $3(u')^4-2(v')^4=-2\Rightarrow u'=$even. Hence $u'=2u''$ and $3\cdot 16 (u'')^4-2(v')^4=-2\Rightarrow 24(u'')^4-(v')^4=-1\Rightarrow$
$$ 24(u'')^4+1=((v')^2)^2. $$ But from the minimality of $y=2s+1$, we have $y\leq (v')^2$. However we have assumed that $s=2s_1$ and $3|s_1$. Then we have written $s_1=3s_2$ $s=6s_2$, $s_2=u^4$, $6s_2+1=v^4$, $v^2=2v_1+1$, $u=u_1=5k$. Also we assumed that $v_1=$even and $3|v_1$, $v_1=6v_2$ and $v'=6v_2+1$. Hence $y=12s_2+1=2(v^4-1)+1=2v^4-1=2(v^2)^2-1=2(2v_1+1)^2-1=2(12v_2+1)^2-1=2(2(v'-1)+1)^2-1=2(2v'-1)^2-1>v'$, when $v'\geq2$. If $v'=1$, then $u''=u'=0\Rightarrow v_2=0\Rightarrow v_1=u_1=0\Rightarrow v=1$. Hence $x=0\Rightarrow y=1$. Which is a solution. However if $v'\geq 2$ we have contradiction. Hence if $v_1\neq 0(\textrm{mod}3)$, we go to the case $v_1+1\equiv 0(\textrm{mod}3)$, $v_1=2(\textrm{mod}3)\Rightarrow v^2=2v_1+1\equiv2(\textrm{mod}3)$, which is not true.

Hence either $v_1=$odd or $s_1\neq 0(\textrm{mod}3)$ and in this last assumption the case $(i)$ is not valid.

Hence we left to examine $v_1=$odd. In this case $(v_1,2v_1+2)=1$ and if $3|v_1$, we have from (1) $v_1=6v_2+3$ and $(6v_2+3)(6v_2+4)=3u_1^4\Leftrightarrow 2(2v_2+1)(3v_2+2)=u_1^4$. But $(2v_2+1,6v_2+4)=1$. Hence exist $u',v'$ such that $(u')^4=2v_2+1$ and $(v')^4=6v_2+4$. Hence $3(u')^4-(v')^4=-1\Leftrightarrow 3(u')^4+1=(v')^4$, with $v'=$even. Set $v'=2v''$. Then $3(u')^4+1=16(v'')^4:(eq)$. Hence $u'=2u''+1=$odd and $5|u'$. Hence $u'=10t+5$. Setting this into $(eq)$ we get $19+75t+75 t^2-4(v'')^4=0$ which is imposible since $t+t^2=$even.

Hence $v_1=$odd and $v_1\neq0(\textrm{mod}3)$. In this case $3|(v_1+1)$ and $v_1=2(\textrm{mod}3)$. Hence $v_1=6v_2+5$ and (1) becomes $(6v_2+5)(4v_2+4)=u_1^4$ or equivalently $(u')^4=6v_2+5$, $(v')^4=4v_2+4$ (that is $(6v_2+5,4v_2+4)=1$). Hence $3(v')^4-2(u')^4=2$ and $2|v'\Rightarrow v'=2v''\Leftrightarrow 3\cdot 8(v'')^4-(u')^4=-1\Leftrightarrow 24(v'')^4+1=(u')^4$. But from the minimality of $y=2s+1$, we have $y\leq (u')^2$. However (assuming $3|s_1$) $s=6s_2$, $v^4=6s_2+1$, $v^2=2v_1+1$, $v_1=6v_2+5$. Hence $v_1=\frac{v^2-1}{2}=(u')^4\Rightarrow v^4=(2(u')^4-1)^2=6s_2+1=2s_1+1=s+1=\frac{y+1}{2}$. Hence when $y\leq (u')^2$, we have $u'=\pm 1$. Hence $v_1$ is not odd.

Thus the case (i) i.e. $3|s_1$ is valid only if $s_1=0$ i.e. when $y=1$.

ii) If $s_1=3s_2+1$, then $3x^4=(3s_2+1)(6s_2+3)\Leftrightarrow x^4=(3s_2+1)(2s_2+1)$. But $(3s_2+1,2s_2+1)=1$. Hence $u^4=3s_2+1$ and $v^4=2s_2+1$. Hence $3v^4-2u^4=1$, $u,v\geq0$ : $(eq)$. Using $(eq)$ we have: $y^2=24x^4+1=24(3s_2+1)(2s_2+1)+1=144s_2^2+120s_2+25=16u^8+12v^4-3=16u^8+8u^4+1=(4u^4+1)^2$. Hence $y=4u^4+1$. If $u\equiv0(\textrm{mod}5)$, then $5|x$ and from $24x^4+1=y^2$, we get $y\equiv1(\textrm{mod}5)$. But from $16u^8+12v^4-3-y^2\equiv 0(\textrm{mod}5)$, we get contradiction. Hence $u\neq 0(\textrm{mod}5)$. But then $y=5y_1$ and $x\neq0(\textrm{mod}5)$ and $s_2=5s_3\Rightarrow 25+600 s_3+3600 s_3^2=25y_1^2\Rightarrow y_1=12 s_3+1$. Hence $y=5(12s_3+1)\equiv 2(\textrm{mod}3)$. Hence $y^2\equiv1(\textrm{mod}3)\Rightarrow 3|x$. Hence $3|v\Rightarrow v=3v_1$, $v^4=2s_2+1$ and we lead to $s_2=40+405t$. Hence for to hold $y^2=24x^4+1$ we must have $$ (1+10t)(121+1215t)=x_1^4\textrm{, }x=3x_1. $$ But $(1+10t,121+1215t)=1$. Hence setting $1+10t=k^4$ and $121+1215t=l^4$, we arrive to $$ 3^5k^4-2l^4=1.\tag 2 $$ When (2) have solution $(k,l)$, then exist $(x',y')$ such that $24(x')^4+1=(y')^2$, with $(x')^4=81k^4l^4$, $y'=4l^4+1$. And if $(x,y)$ is the minimal solution of $24x^4+1=y^2$, then $x^4=81u^4v_1^4$, $y=4u^4+1$ and it must be $y<y'\Leftrightarrow u^4<l^4\Leftrightarrow \frac{1}{2}(-1+243k^4)<\frac{1}{2}(-1+243k^4)$, which is not true. Hence $y=y'$, which is contradiction from the minimality of $y$. Hence the equation (2) have no solutions.

iii) If $s_1=3s_2+2$, then $3x^4=(3s_2+2)(6s_2+5)$, which is imposible.

2) If $s=$odd$=2s_1+1$, then $6x^4=(2s_1+1)(2s_1+2)\Leftrightarrow 3x^4=(s_1+1)(2s_1+1)$.

i) If $3|(s_1+1)\Rightarrow s_1+1=3s_2\Rightarrow$ $x^4=s_2(6s_2-1)$ and $(s_2,6s_2-1)=1$. Hence $u^4=s_2$ and $v^4=6s_2-1$, which is imposible.

ii) If $s_1+1=3s_2+1$. Hence $3x^4=(3s_2+1)(6s_2+1)$, which is again imposible.

iii) If $s_1+1=3s_2+2\Rightarrow s_1=3s_2+1$. Hence $3x^4=(3s_2+2)(6s_2+3)\Rightarrow$ $x^4=(3s_2+2)(2s_2+1)$. But $(3s_2+2,2s_2+1)=1$. Hence $u^4=3s_2+2$ and $v^4=2s_2+1$, which is again imposible ($u\equiv0,1,2(\textrm{mod}3)\Rightarrow u^4\equiv0,1(\textrm{mod}3)$).