For $x^2-3y^2=1$ over integers more than 1, can $\frac{y+1}2$ be square number?
I know that $x^2-3y^2=1$ is one of pell's equation, so I know its general solution. But I know nothing about its properties, and I can't proceed my proof. How should I approach this question?
On
Of course this is equivalent to $y = 2z^2 - 1$ for some $z$.
Recall that all solutions to this Pell equation are parametrized by $x+y\sqrt 3 = (2 + \sqrt 3)^n$. Looking modulo 2, we see that we must have $n = 2k+1$ for some $k$. Denoting $a + b \sqrt 3 = (2 + \sqrt 3 )^k$, we see that $x + y\sqrt 3 = (a^2 + 3b^2 + 2ab \sqrt 3 )(2 + \sqrt 3)$, so we need to solve the equation
$2z^2 - 1 = a^2 + 3b^2 + 4ab$, where we have $a^2 - 3b^2 = 1$. Substituting this, we get $2z^2 - 1 = 2a^2 + 4ab - 1$, that is $z^2 = a(a+2b)$.
$\gcd(a, a+2b) = \gcd(a,2)$, so we split into cases.
Case 1 $\gcd(a,2)=1$. In this case $a$ must be square, so it is sufficient to solve the equation $a^4 - 3b^2 = 1$. We will return to this later.
Case 2 $\gcd(a,2)=2$. Then $a+2b = 2m^2$ for some $m$. Substituting this we get $(2m^2 - 2b)^2 - 3b^2 = b^2 - 8m^2b + 4m^4 = 1$ or
$b^2 - 8m^2b + (4m^4 - 1) = 0$
This is a quadratic equation in $b$, so in order for it to have integer solutions, the discriminant must be a square, that is $12m^4 + 1 = \ell^2$ for some $\ell$.
$\ell^2 - 3(2m^2)^2 = 1$, so $\ell + 2m^2 \sqrt 3 = (2 + \sqrt 3)^2t$ (we know that it is an even power because modulo 2). Denoting $c + d \sqrt 3 = (2 + \sqrt 3)^t$ we get $\ell^2 = cd$, so in particular $c$ is a square, and once again we find that it is sufficient to solve the equation $a^4 - 3b^2 = 1$.
Very cool lemma: Defining $x_n + y_n \sqrt d = (x_1 + y_1 \sqrt d)^n$, where $x_1 + y_1 \sqrt d$ is a solution to the Pell equation for $d$, we always have that $x_{4k}$ is never a square.
Proof: You can easily check that $x_{4k} = 8x_k^4 - 8x_k^2 + 1$, so it suffices to prove that the equation $y^2 = 8x^4 - 8x^2 + 1$ has no solutions.
$y^2 - 2(2x^2 - 1)^2 = -1$, and by the same technique as before (defining $a + b\sqrt 2 = (1 + \sqrt 2)^k$ we find that $2x^2 - 1 = a^2 + 2b^2 + 2ab$ where $a^2 - 2b^2 = +- 1$. We divide into cases:
Case 1: $a^2 - 2b^2 = 1$. We then get $2x^2 - 1 = 2a^2 + 2ab - 1$, that is $x^2 = a(a+b)$. In particular, $a$ is a square so we get the equation $a^4 - 2b^2 = 1$. This is well known to have only trivial solutions, my favorite proof moving things over and squaring we get $(\frac{a^4 + 1}{2})^2 - a^4 = b^4$ which is a Pythagorean triple containing two squares, and that is known since Fermat to have only trivial solutions.
Case 2: $a^2 - 2b^2 = -1$. We then get $2x^2 - 1 = 4b^2 + 2ab -1$, that is $x^2 = b(a+2b)$. In particular, $a+2b$ is a square (trying to use $b$ here is a trap: the equation $y^2 - 2x^4 = -1$ is notoriously hard and has the solution $y=239, x=13$). Setting $a = k^2 - 2b$ and substituting we get
$k^4 - 4bk^2 + 4b^2 - 2b^2 = -1$
$2b^2 - 4bk^2 + (k^4 + 1)$. This is a quadratic in $b$, so the discriminant must be a square, that is $2(k^4 - 1)$ is a square, or $k^4 - 1$ is twice a square, but we have just solved that.
So the lemma is proved.
Now in our case we have $x_n + y_n \sqrt 3 = (2 + \sqrt 3)^n$, and we want to show that $x_n$ is not a square for $n > 0$. From the lemma we just proved, we know that $n$ is not divisible by 4. Notice that if $x_n = k^2$ is even, then $x_n^2$ is divisible by 8 and we get $-3y_n^2 \equiv 1 \mod 8$ which is impossible. Therefore $x_n$ is odd and so $n$ must be even, so $n = 4\ell + 2$. However, let's look at $(2 + \sqrt 3)^n$ modulo 4.
$(2 + \sqrt 3)^2 = 7 + 4\sqrt 3 \equiv -1 \mod 8$ and so $(2 + \sqrt 3)^{4\ell + 2} \equiv (-1)^{2\ell + 1} = -1 \mod 4$, but squares are never $-1 \mod 4$, which is a contradiction.
QED
In fact, a more general result is true: If $x^4 - Dy^2 = 1$, then $x^2 + y \sqrt D$ is either the first or the second smallest solution to the relevant Pell equation, and the only time both of them are squares is $D = 1785$. This was proved by J.H.E. Cohn (once I recall the paper I will add it in the comments) using a rather heavy result of Ljunggren and some ingenious computations with Jacobi symbols.
On
$$x^2-3y^2=1\implies y^2=\frac{x^2-1}{3}\implies \frac{x-1}{p}\cdot\frac{x+1}{q}\quad\text{where}\quad p,q\quad \text{ divides }\quad 3$$
It is easy to see the solutions of $p=1,q=3\text{ and }x=1\lor x=2$ but perhaps there are other values of x divisible by these factors and, it happens there are.
For $x-1$, x can be any integers and, $x+1$ can be any multiple of $3$ such as $3,6,9$ but the result, divided by 3 must be a perfect square and these get rarer with altitude. Here is a sample of infinite $(x,y)$ solutions. Only positive integers are shown for simplicity but negatives apply as well.
$$(x,y)\in\{(1,0),(2,1),(7,4),(26,15),(97,56),(362,209), \cdots\}$$ This gives no definition of the set. A search is still required and although a solution for $x$ would be faster it alone yields just few insights into the values of $y$ that yield integers.
$$x^2-3y^2=1\implies x^2={3y^2+1}$$ There is a brighter side though, in that both $x$ and $y$ values are known sequences in the Online Encyclopedia of Integer Sequences.
Sequence A001075 shows $x\in\{ 1, 2, 7, 26, 97, 362, \cdots\}$
Sequence A001353 shows $y\in\{ 0, 1, 4, 15, 56, 209, \cdots\}$
These sequences often come with several formulas for their generation and perhaps one of them may meet your needs in generating the $n^{th}$ pair directly.
ADDED: kind of neat, the numbers of interest factor as the product of two sequences, both staisfy $$ w_{n+2} = 4 w_{n+1} - w_n \; , \; $$ one sequence begins $ 2, 7, 26, 97..$ and the other starts $1, 4, 15, 56, ...$ It is therefore possible this could be finished without elliptic curves. ADDED EXTRA: Alright, worth investigating, the two sequences are precisely the pairs of solutions to $u^2 - 3 v^2 = 1,$ as in $2^2 - 3 \cdot 1^2 = 1 \;, \; \; $ $7^2 - 3 \cdot 4^2 = 1 \;, \; \; $ $26^2 - 3 \cdot 15^2 = 1 \;, \; \; $ $97^2 - 3 \cdot 56^2 = 1 \;, \; \; $ There is, though, a bit of an index shift: we are not using $15 \cdot 26,$ we are using $15 \cdot 7.$
==============
ORIGINAL
Just curious. Mordell's book points out that there are just finitely many integer solutions to $$ u^2 = a v^4 + b v^2 + c, $$ for example pages 236, 257, 268. The suitable values $y_n,$ the odd ones, satisfy $$ y_{n+2} = 14 y_{n+1} - y_n $$ beginning with $y_1 = 1$ and $y_2 = 15$ then $y_3 = 209.$