# Are there positive integers $a$ and $b$ such that $a(a^2-1)=2b^2$? [closed]

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

$$a(a^2-1)=2b^2$$

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

$$a(a^2-1)=a(a-1)(a+1)$$. 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 $$a \in [-1,0] \cup [1,+\infty[$$

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