# Math Challenge - March 2020

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Infrared
Gold Member
I still don't think it's right. It looks like to me that you're claiming that since $3$ divides $8q^2-p^2$, it must divide both $8q^2$ and $p^2.$ I don't see why this should be true without further argument. For example, $3$ divides $1+5$ but not $1$ or $5$.

PeroK
Homework Helper
Gold Member
Generalisation:
Let's look at $f(x) = x^3 - px$, where $p$ is prime.

We'll look for $f(x) = f(y)$, where $x = \frac a b, \ y = \frac c d$ are in the lowest form.
$$f(\frac a b) = f(\frac c d) \ \Rightarrow \ \frac{a(a^2 - pb^2)}{b^3} = \frac{c(c^2 - pd^2)}{d^3}$$

Note that $\frac{a(a^2 - pb^2)}{b^3}$ is also in its lowest form, so we have $b = d$. And:
$$a(a^2 - pb^2) = c(c^2 - pb^2) \ \Rightarrow \ a^2 + ac + c^2 = pb^2 \ \Rightarrow \ (2a+c)^2 = 4pb^2 - 3c^2$$
So, we need solutions to:
$$4pb^2 - n^2 = 3c^2$$
Where $n = 2a + c$.

We have solutions for $p = 3$. Namely, $x = -2, y = 1$ and $x = -1, y = 2$.

Otherwise, note that neither of $b, n$ can be divisible by $3$ without $b, c$ having a common factor of $3$. Hence $b^2 = n^2 = 1 \ (mod \ 3)$ and we have:
$$p = 1 \ (mod \ 3)$$
For example, we have no solutions for $p = 2, 5$, but we have a triple solution for $p = 7$:
$$x= -3, y = 1, z = 2\ \ \text{with} \ f(x) = f(y) = f(z) = -6$$

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