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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?

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  • $\begingroup$ 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 $3$. $\endgroup$ – Bart Michels Nov 23 '14 at 21:13
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    $\begingroup$ 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 '16 at 9:13
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    $\begingroup$ @Supersingularity, good find. And the answer there actually treats the problem like it's trivial, whithout solving it $\endgroup$ – Yuriy S Apr 15 '16 at 17:39
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    $\begingroup$ 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 $a=b=c= \pm 1$. $\endgroup$ – Supersingularity Apr 15 '16 at 20:43
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    $\begingroup$ $2a^2+1 = 3c^2$ is not a Pell equation! $\endgroup$ – Piquito Apr 20 '16 at 1:58
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${\bf Summary:}$ 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. $\Box$

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    $\begingroup$ 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 '16 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. $\endgroup$ – Kieren MacMillan Apr 21 '16 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. $\endgroup$ – Kieren MacMillan Apr 21 '16 at 13:28
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    $\begingroup$ @san, math.rice.edu/~evanmb/math499spring10/vigrenotes3.pdf. $\endgroup$ – zhoraster Apr 21 '16 at 14:14
  • $\begingroup$ @zhoraster: Well, with that parameterization, plus san's last lemma, there's the elementary proof I wanted. $\endgroup$ – Kieren MacMillan Apr 21 '16 at 15:03
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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 $r=2s+3p$.

[PROOF INCORRECT AFTER THIS POINT]

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.

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    $\begingroup$ 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 '16 at 8:22
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    $\begingroup$ 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 '16 at 3:26

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