Proving that a list of perfect square numbers is complete

elliptic-curvesnumber theoryproof-writingsquare-numbers

Well, I have a number $n$ that is given by:

$$n=1+12x^2\left(1+x\right)\tag1$$

I want to find $x\in\mathbb{Z}$ such that $n$ is a perfect square.

I found the following solutions:

$$\left(x,n\right)=\left\{\left(-1,1^2\right),\left(0,1^2\right),\left(1,5^2\right),\left(4,31^2\right),\left(6,55^2\right)\right\}\tag2$$

Is there a way to prove that this a complete set of solutions? So I mean that the solutions given in formula $(2)$ are the only ones?

My work:

We know that:
$$
1 + 12x^2 \left(1+x \right) \ge 0
\space \Longleftrightarrow \space
x \ge -\frac{1+2^{-2/3}+2^{2/3}}{3}
\approx -1.07245
\tag3
$$
So we know that for $x<-1$ there are definitely no solutions.

Best Answer

$y^2=1+12x^2(1+x) \implies (12 y)^2 = (12 x)^3 + 12 (12 x)^2 + 144$

Magma code for positive $y$ only:

S:= IntegralPoints(EllipticCurve([0,12,0,0,144]));
for s in S do
  x:= s[1]/12;
  if x eq Floor(x) then
    print "(",x,", ",Abs(s[2]/12),")";
  end if;
end for;

Output:

( -1 ,  1 )
( 0 ,  1 )
( 1 ,  5 )
( 4 ,  31 )
( 6 ,  55 )

Related Solutions

Find All Integers n Such That 2^{n-1}*n+1 is a Perfect Square – Number Theory

Suppose $2^{n-1} n + 1 = x^2$ with $n > 5$. Then $2^{n-1} n = (x-1)(x+1)$ is a product of two integers that differ by $2$. Corresponding to this we must have $x-1 = 2^k u$ and $x+1 = 2^{n-1-k} v$ where $uv = n$. Now $\min(k,n-1-k) \le 2$, otherwise $x+1$ and $x-1$ would differ by at least $4$. If $k \le 2$, $x+1 \ge 2^{n-3}$ and $x-1 \ge 2^{n-3}-2$ so $$2^{n-1} n = (x+1)(x-1) \ge 2^{2n-6} - 2^{n-2} $$ which simplifies to $$ n + 1/2 \ge 2^{n-5}$$ and it is easy to see that this is false if $n \ge 9$. Similarly if $n-1-k \le 2$, $ x-1 \ge 2^k \ge 2^{n-3}$ and $x+1 \ge 2^{n-3}+2$ so $$ 2^{n-1} n \ge 2^{2n-6} + 2^{n-2}$$ and this is false if $n$ is too big.

Proving a statement about perfect squares

Given a constant $\, D := 108,\,$ consider the function defined by $$ f(x,y) := 9 + D\,x^2(1+x) - y^2 \tag{1} $$ where $\, f(x,y) = 0\,$ is the equation of an elliptic curve. It is equivalent to the curve in Weierstrass form $$ E: y^2 = x^3 + 1/D\,x^2 + 12/D^3 \tag{2} $$ in the following sense. $\,f(x,y) = 0 \,$ iff $\,[x/D,y/D^2]\,$ is a point of curve $\,E\,$ which has $j$-invariant equal to $\,-12288/25.\,$ This curve is equivalent to the LMFDB 135.a1 curve $$ E135a: y^2 + y = x^3 - 3x + 4. \tag{3} $$ The curve $\,E\,$ has a rational point $\,P:=[1/D,15/D^2]\,$ of rank $\,1\,$ which PARI/GP seems not to be able to provide, but it does provide a rational generating point $\,[4,-8]\,$ for curve $\,E135a\,$. Given a point $\,[x,y]\,$ of $\,E135a\,$ then $\,[(x-1)/(3D),(-2y-1)/D^2]\,$ is a point of $\,E.\,$ Each point $\,[x,y]\,$ of $\,E\,$ satisfies $$ 0 = f(x\, D,y\, D^2). \tag{4} $$ However there are only a finite number of integer solutions to $\,f(x,y)=0.\,$ They correspond to the generator multiples $$ 1P\mapsto(1,15),\, 2P\mapsto(0,3),\, 4P\mapsto(-1,-3),\, 7P\mapsto(6,-135),\, 8P\mapsto(4,93). \tag{5}$$ Each solution $\,(x,y)\,$ yields a solution $\,(x,-y).\,$ There are no other solutions in integers.

Some PARI/GP code is

D = 108; E = ellinit([0,1/D,0,0,1/12/D^2]); P = [1/D,15/D^2];
E135a = ellinit([0, 0, 1, -3, 4]);
print("E:",ellgenerators(E));
print("E135a:",ellgenerators(E135a));
for(n=1, 9, print(n," ",[x,y]=ellmul(E,P,n); [x*D,y*D^2]));

Best Answer

Related Solutions

Find All Integers n Such That 2^{n-1}*n+1 is a Perfect Square – Number Theory

Proving a statement about perfect squares

Related Question