Let $a,b,c$ be positive integers satisfying \begin{align} 2a^2-1 &= b^2, \\ 2a^2+1 &= 3c^2. \end{align} The trivial solution is $(a,b,c)=(1,1,1)$. Are there others?

  • $\begingroup$ It is equivalent to $(2a^2-1)(2a^2+1)=3d^2$, that is, $(2a^2)^2-3d^2=1$. By Pell-theory, $2a^2+d\sqrt3=(2+\sqrt3)^n$ for some $n>0$ but I'm not sure how to proceed from here. Certainly $n$ is odd by inspection mod $3$. $\endgroup$ Nov 23, 2014 at 21:13
  • 3
    $\begingroup$ Continuing @barto's comment, one needs to show that $(2+\sqrt{3})^n + (2-\sqrt{3})^n = \lceil (2+\sqrt{3})^n\rceil$ is not a square. Since the residues are cyclic, and $4$ is already there, any modular attempts will fail. An interesting question, would be nice to see a solution. $\endgroup$
    – zhoraster
    Apr 13, 2016 at 9:13
  • 1
    $\begingroup$ @Supersingularity, good find. And the answer there actually treats the problem like it's trivial, whithout solving it $\endgroup$
    – Yuriy S
    Apr 15, 2016 at 17:39
  • 1
    $\begingroup$ Indeed that is frustrating. Here is something else someone might be able to do better with: we have $b+a\sqrt{2}$ is a unit in $\mathbb{Z}[\sqrt{2}]$, so it looks like $\pm (1+\sqrt{2})^n$ for some $n$ (really more $\pm$'s but I'm just simplifying. Hence $b = \sum_{k=0}^{n/2} \binom{n}{2k} 2^k$ for some $n$. On the other hand, because $b + c\sqrt{3}$ lies over 2 in $\mathbb{Z}[\sqrt{3}]$, we should have $b+c\sqrt{3} = (\sqrt{3}+1)(2+\sqrt{3})^n$ (something like this). Expanding again, we get another binomial type expression for $b$. Maybe comparing them would force $n$ to give $a=b=c= \pm 1$. $\endgroup$ Apr 15, 2016 at 20:43
  • 1
    $\begingroup$ $2a^2+1 = 3c^2$ is not a Pell equation! $\endgroup$
    – Piquito
    Apr 20, 2016 at 1:58

2 Answers 2


${\bf Summary:}$ We prove that for any positive integer solution $a$, $b$ and $c$ to the given equations there exist $x,y\in \Bbb{Z}$ such that $$ 4a=x^2+3y^2,\quad b=\frac{|x^2-3y^2|}{2}\quad\text{and}\quad c=|xy|. $$ We insert these values into the second equation and obtain $(3(x^2-y^2))^2=8(x^4-1)$ which implies $x^2=y^2=1$, hence $a=b=c=1$.

${\bf Edit:}$ Thanks to a hint of Zhoraster to the rational slope method, the first part has been simplified.

Assume $a$, $b$ and $c$ are an integer solution to the given equations. Adding the two equations we obtain: $$ 4a^2=b^2+3c^2\quad\Rightarrow\quad 1=\left(\frac{b}{2a}\right)^2+3\left(\frac{c}{2a}\right)^2. $$ So $\left(\frac{b}{2a},\frac c{2a}\right)$ is a rational point of the ellipse $$ 1=u^2+3v^2. $$ If $u,v\in\Bbb{Q}$, then the slope of the line through $(u,v)$ and $(1,0)$ is rational, i.e., there exists $t\in\Bbb{Q}$ with $t(u-1)=v$. But then, if $(u,v)$ is on the ellipse, $$ 1=u^2+3(t(u-1))^2\quad \Rightarrow\quad (u-1)(3t^2(u-1)+u+1)=0. $$ If $u\ne 1$, we obtain $$ u=\frac{3t^2-1}{3t^2+1}\quad\text{and}\quad v=t(u-1)=-\frac{2t}{3t^2+1}, $$ so all rational points of the ellipse different from $(1,0)$ are parametrized by $t\in \Bbb{Q}$. Write $t=\frac yx$ for some coprime integers $x,y$, then $$ u=\frac{3y^2-x^2}{3y^2+x^2}\quad\text{and}\quad v=t(u-1)=-\frac{2xy}{3y^2+x^2}. $$ In particular, for $(u,v)=\left(\frac{b}{2a},\frac c{2a}\right)\ne (1,0)$ there exist $x,y$ coprime integers, such that $$ \frac{b}{2a}=\frac{3y^2-x^2}{3y^2+x^2}\quad\text{and}\quad \frac{c}{2a}=-\frac{2xy}{3y^2+x^2}. $$ Now $x,y$ cannot be both even, since they are coprime, and if one is even and the other is odd, then $3y^2-x^2$ and $3y^2+x^2$ are both odd, which contradicts $\frac{b}{2a}=\frac{3y^2-x^2}{3y^2+x^2}$, since $2\not|b$. So $x$ and $y$ are both odd. Moreover, if $3|x$, then we can set $x_1=y$, $y_1=\frac x3$, and then $x_1, y_1$ are coprime, $3\not| x_1$ and $$ \frac{b}{2a}=\frac{x_1^2-3y_1^2}{3y_1^2+x_1^2}\quad\text{and}\quad \frac{c}{2a}=-\frac{2x_1y_1}{3y_1^2+x_1^2}. $$

So, we have achieved the following result: If $a$, $b$ and $c$ are integer solutions to the given equations, there exists $x,y$ coprime odd integers, such that $3\not| x$ and such that $$ \frac{b}{2a}=\pm\frac{3y^2-x^2}{3y^2+x^2}\quad\text{and}\quad \frac{c}{2a}=-\frac{2xy}{3y^2+x^2}. $$ Since $gcd((3y^2+x^2)/2,xy)=1=gcd(c,2a)$, we have

$$ c=\pm xy, \quad \pm 4a=3y^2+x^2\quad\text{and}\quad b=\pm\frac{3y^2-x^2}2. $$

Now we insert these values into the second equation $2a^2+1=3c^2$ (or in the equivalent equation $16a^2+8=24 c^2$) and obtain $$ (x^2+3y^2)^2+8=24x^2y^2\quad \Leftrightarrow \quad (3(x^2-y^2))^2=8(x^4-1). $$ By the Lemma below, in $\Bbb{Z}$ the solutions of this equation satisfy $x^2=1$ and $x^2=y^2$, hence $(a,b,c)=(1,1,1)$ is the only positive integer solution.

${\bf Lemma:}$ The only two integer solutions of $R^2=8(x^4-1)$ satisfy $R=0$ and $x^2=1$.

${\bf Proof:}$ Evidently $R=0$ and $x^2=1$ give solutions. Assume by contradiction that $x\ne \pm 1$. We also can assume that $x>0$. We have $$ R^2=2^3(x^2+1)(x+1)(x-1), $$ and clearly $x$ is odd (else $2^3$ would be a square). Since $$ gcd(x^2+1,x+1)=gcd(x^2+1,x-1)=gcd(x+1,x-1)=2, $$ any prime $p>2$ which divides $x^2+1$ cannot divide $x+1$ or $x-1$, hence $x^2+1=2^i r^2$ for some $i,r\ge 1$, with $r$ odd. Similarly $x+1=2^js^2$ and $x-1=2^k t^2$ for some $j,k,s,t\ge 1$, with $s,t$ odd. Moreover, since $2^{i+j+k+3}$ is a square, necessarily $i+j+k$ is odd. But, if $i$ is even, then $x^2+1=(2^{i/2}r)^2$, which is impossible, since $x\ne 0$. So $i$ is odd and $j+k$ is even.

We have $$ 2=x+1-(x-1)=2^j s^2-2^k t^2\quad \Rightarrow\quad 1=2^{j-1} s^2-2^{k-1} t^2, $$ and so $j=1$ or $k=1$, hence both $j$ and $k$ are odd.

But then $1=(2^{(j-1)/2}s)^2-(2^{(k-1)/2}t)^2$, which is impossible, since $t>0$.

This contradiction concludes the proof. $\Box$

  • 2
    $\begingroup$ In fact, the part until the last lemma is proved easier. One may add the equations, yielding $4a^2 = b^2 + 3c^2$. The last equation is solved in a standard way by considering rational secants of $x^2 + 3y^2 = 4$, yielding exactly your first claim. $\endgroup$
    – zhoraster
    Apr 21, 2016 at 6:25
  • $\begingroup$ Seems nice! I'm going to let it sit for a day or two while I (and others) look it over, and then accept if no flags are raised. $\endgroup$ Apr 21, 2016 at 13:28
  • $\begingroup$ Would still be nice to find an elementary solution, of course — but that won't stop me from accepting this answer. $\endgroup$ Apr 21, 2016 at 13:28
  • 1
    $\begingroup$ @san, math.rice.edu/~evanmb/math499spring10/vigrenotes3.pdf. $\endgroup$
    – zhoraster
    Apr 21, 2016 at 14:14
  • 1
    $\begingroup$ @zhoraster: Well, with that parameterization, plus san's last lemma, there's the elementary proof I wanted. $\endgroup$ Apr 21, 2016 at 15:03

NOTE: As pointed out by Erick Wong, there was a flaw in the original proof. It may be fixable, though I haven't found a fix yet; leaving here for history (and just in case anyone else can take this method to the goal line).

Theorem. The Diophantine system of equations \begin{align} 2a^2-1 &= b^2, \\ 2a^2+1 &= 3c^2 \end{align} has only the trivial solution $\lvert a\rvert = \lvert b \rvert = \lvert c \rvert = 1$.

Proof. Evidently, $\lvert a\rvert = \lvert b \rvert = \lvert c \rvert = 1$ is a solution. We also see that $\lvert c \rvert = 1$ forces $\lvert a\rvert = \lvert b \rvert = \lvert c \rvert = 1$. We now assume, contrary to the theorem, that there is a solution with $bc > 1$ [positivity assumed without loss of generality].

Multiplying the two original equations yields \begin{align} (2a^2-1)(2a^2+1) &= (b^2)(3c^2) \\ (2a^2)^2 - 1 &= 3(bc)^2. \end{align} Evidently $bc$ is odd, say $bc=2u+1$ for an integer $u \ge 1$ [because $bc > 1$]. There are exactly two cases to consider.

Case 1: $a \equiv 1\!\pmod{3}$. Say $a=3v+1$ for an integer $v \ge 1$ [because $a > 1$]. Substituting, expanding, and simplifying yields \begin{align} \bigl(2(3v+1)^2\bigr)^2 - 1 &= 3(2u+1)^2 \\ 324v^4 + 432v^3 + 216v^2 + 48v + 3 &= 12u^2+12u+3 \\ v(3v+2)(9v^2+6v+2) &= u(u+1). \end{align} Since all the factors are positive [by hypothesis], the Fundamental Theorem of Arithmetic implies there exist positive integers $p,q,r,s$ such that \begin{align} \tag{1} v(3v+2) &= pq, & u &= pr, \\ 9v^2+6v+2 &= rs, & u+1 &= qs. \end{align} The left-hand pair of relations implies $rs-3pq=2$, and the right-hand pair implies $qs-pr=1$. Solving these two equations for $q$ and $r$ in terms of $s$ and $p$ yields \begin{align} \tag{2} q &= \frac{s+2p}{s^2-3p^2}, & r &= \frac{2s+3p}{s^2-3p^2}. \end{align} Since $q$ and $r$ are both integers [by hypothesis], it must be that $s^2-3p^2$ divides both $s+2p$ and $2s+3p = 2(s+2p)-p$, and so it must divide $p$. But $qs-pr=1$ implies $\gcd(p,s)=1$, and hence $\gcd(s^2-3p^2,p)=1$. Therefore we conclude $s^2-3p^2 = \pm 1$; since $s^2-3p^2=-1$ has no integer solutions, we have $s^2-3p^2=1$. Substituting this into (2) gives $q=s+2p$ and $r=2s+3p$.


NOTE: One can quickly deduce that $(s,p)$ and $(r,q)$ are consecutive solutions to $U^2-3V^2=1$, with $r>s$ and $q>p$. Since $(2a^2,bc)$ is also a solution, we can let $(s,p)$ be the $n$th solution, so that $(r,q)$ is the $(n+1)$th, and $(2a^2,bc)$ is the $(2n+1)$th. One can then show [using standard elementary Pell theory] that $(b,c)=(q+p,s+p)$, so that $b-c=2p$. From that point, I can't get any more useful information with this method.

  • 1
    $\begingroup$ From the last system I only obtain $(3p^2-s^2)(3q^2-r^2)=1$. How you obtain $qs=1$? $\endgroup$
    – san
    Apr 19, 2016 at 8:22
  • 2
    $\begingroup$ Sorry, there's a typo in the line where $r-q$ is first calculated: $r-q = s+p$, not $s-p$. This seems to eliminate the contradiction of $(s-p)^2 = s^2 - p^2$. $\endgroup$
    – Erick Wong
    Apr 20, 2016 at 3:26

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.