Math Challenge - March 2020

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  • #76
Infrared
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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##.

Also, in the future, please write your solutions in your post- I'd rather not have to follow links.
 
  • #77
PeroK
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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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