If $a$ and $b$ are positive integers, does a solution to the follwing equation exist?


I have tried graphing to no avail, can anybody help?


4 Answers 4


$a(a^2-1)=a(a-1)(a+1)$. The three factors are consecutive integers and hence all coprime with the possible exception of a single factor of $2$ if $(a-1)$ and $(a+1)$ are both even. In order for three coprime factors to multiply to $2b^2$, two of the factors must be squares (the third must be twice a square). But the factors differ from each other by either $1$ or $2$, and there are no integer squares other than $0,1$ that differ from each other by either $1$ or $2$. So one solution is: $a-1=0$, $a=1$, $a+1=2$ yielding $a(a^2-1)=a(a-1)(a+1)=0$ corresponding to $b=0$. There are no other integer solutions.


There is no positive integer solution to the equation $$a(a+1)(a-1) = 2b^2$$

Assume the contrary, let's say $(a,b)$ is a positive integer solution.

Since $\gcd(a,a^2-1) = 1$, we can find $c,d \in \mathbb{Z}_{+}$ such that $b = cd$ and one of the following is true:

$$(a,a^2-1) = (2c^2,d^2)\quad\text{ or }\quad (a,a^2-1) = (c^2,2d^2)$$

We can rule out the first case because it implies $a^2 - d^2 = 1$ and we know the difference between two non-zero perfect square is never $1$.

This means $a = c^2$ and $(a+1)(a-1) = 2d^2$.

Since $2d^2$ is even, $a$ is odd and both $a+1$, $a-1$ is even. This forces $d$ to be even.

Let $d = 2e$. We have $\left(\frac{a+1}{2}\right)\left(\frac{a-1}{2}\right) = 2e^2$.

Since $\gcd\left(\frac{a+1}{2}, \frac{a-1}{2}\right) = 1$, we can find $f, g \in \mathbb{Z}_{+}$ such that $e = fg$ and one of the following is true:

$$\left(\frac{a+1}{2}, \frac{a-1}{2}\right) = (2f^2,g^2)\quad\text{ or }\quad \left(\frac{a+1}{2}, \frac{a-1}{2}\right) = (f^2,2g^2)$$

We can rule out the second case because it implies $c^2 - (2g)^2 = 1$ which is impossible.

This leaves us with $\begin{cases} c^2 + 1 = a + 1 = 4f^2\\ c^2 - 1 = a - 1 = 2g^2 \end{cases}$.

Since RHS is even, we find $c$ is odd. However, we know for any odd integer $n$, $n^2 \equiv 1\pmod 8$. The first equation $c^2 + 1 = 4f^2$ leads to $2 \equiv 0 \pmod 4$ and this is absurd.

As a result, we can conclude there is no positive integer solution for the equation.


Nice try with graphing, but the problem with this is that you are assuming $a=b$ as you were solving two separate function in $x$.

You can rewrite the equation as:

$$ b=\frac{\sqrt{a}\sqrt{a^2-1}}{\sqrt{2}} $$

Note that one of $a$ or $a^2-1$ will always be even, by assumption.

Let $a=2k$ then $a^2-1= 4k^2-1$ this would easily be factorised and you would be easily seeing that it can never be perfect square for perfect square $k$.

Similarly $a=2k+1$, is easy proof.

If you try, upload this in comment and let me see for mistakes.


$b=±\frac{\sqrt{a(a^2 -1)}}{\sqrt 2}$

You just put the value of $a$ to get the value of $b$. And you can put any value of $a$ such that $$a(a^2-1) \ge 0$$

That is when $a \in [-1,0] \cup [1,+\infty[$

  • 3
    $\begingroup$ This does not answer the question ! $\endgroup$
    – user65203
    Nov 9, 2018 at 15:59
  • $\begingroup$ Ahh I didn't notice the 'integer' $\endgroup$ Nov 9, 2018 at 16:19

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