# Find $a \in \mathbb N$ such that $x^2+ax-1 = y^2$ has a solution in positive integers

Question: Find all the positive integers $$a$$ such that $$x^2+ax-1 = y^2$$ has a solution in positive integers $$(x,y)$$.

Comments: It's easy to see that this equation rarely has a solution (in the sense that for a fixed $$a$$, $$x^2+ax-1$$ is a perfect square only for finitely many values of $$x$$). In fact, if $$a$$ is even then $$x^2 \le x^2+ax-1 < (x+a/2)^2$$, so $$x^2+ax-1$$ is almost never a perfect square. The problem is that I can't control the interval $$[x^2,(x+a/2)^2)$$ when $$a$$ grows. There's a similar argument for $$a$$ odd.

However, it is possible to find some families of such $$a$$'s. For instance, if $$a$$ is a perfect square, say $$a=k^2$$, then there exists the solution $$(x,y) = (1,k)$$.

In addition, if $$x > 1$$ then its prime power factors are $$2$$ and/or $$p^\alpha$$, where $$p \equiv 1 \pmod 4$$. In fact, if $$p|x$$ then the constraint of the equation modulo $$p$$ yields that $$-1$$ is a square.

Actually, every $$a$$ that is not $$2 \pmod 4$$ yields a solution. These $$a$$ are impossible because then $$x^2 + ax - 1 \equiv x^2 + 2x - 1 \equiv (x - 1)^2 - 2 \pmod 4,$$ which is either $$2$$ or $$3$$ mod $$4$$, but squares are only $$0$$ or $$1$$ mod $$4$$.

For the other $$a$$, I will construct values of $$x$$ that work.

For $$a = 4k$$, take $$x = 2k^2 - 2k + 1 = \frac{(2k - 1)^2 + 1}{2}$$. Then \begin{align*} x^2 + ax - 1 &= (2k^2 - 2k + 1)^2 + 4k(2k^2 - 2k + 1) - 1\\ &= (4k^4 - 8k^3 + 8k^3 - 4k + 1) + (8k^3 - 8k^2 + 4k) - 1\\ &= 4k^4\\ &= (2k^2)^2 \end{align*}

For $$a = 4k + 1$$, take $$x = 4k^2 + 1$$. Then \begin{align*} x^2 + ax - 1 &= (4k^2 + 1)^2 + (4k + 1)(4k^2 + 1) - 1\\ &= (16k^4 + 8k^2 + 1) + (16k^3 + 4k^2 + 4k + 1) - 1\\ &= 16k^4 + 16k^3 + 12k^2 + 4k + 1\\ &= (4k^2 + 2k + 1)^2 \end{align*}

For $$a = 4k + 3$$, take $$x = 4k^2 + 4k + 2$$. Then \begin{align*} x^2 + ax - 1 &= (4k^2 + 4k + 2)^2 + (4k + 3)(4k^2 + 4k + 2) - 1\\ &= (16k^4 + 32k^3 + 32k^2 + 16k + 4) + (16k^3 + 28k^2 + 20k + 6) - 1\\ &= 16k^4 + 48k^3 + 60k^2 + 36k + 9\\ &= (4k^2 + 6k + 3)^2 \end{align*}

$$x^2+ax-1 = y^2 \implies (2 x + a)^2 - (2 y)^2 = 4 + a^2$$

gp-code:

axy()=
{
for(a=1, 100,
S= [];
T= thue('x^2-1, 4+a^2);
for(i=1, #T,
x= (T[i]-a)/2;
y= T[i]/2;
if(x>0 & y>0,
if(x==floor(x) & y==floor(y),
S= concat(S, [[x,y]]);
)
)
);
if(#S, print1(a", "))
)
};


Output:

1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 23, 24, 25, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 53, 55, 56, 57, 59, 60, 61, 63, 64, 65, 67, 68, 69, 71, 72, 73, 75, 76, 77, 79, 80, 81, 83, 84, 85, 87, 88, 89, 91, 92, 93, 95, 96, 97, 99, 100,


I.e. $$a\not\equiv 2\pmod{4}$$