- #76

Infrared

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

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- #76

Infrared

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

- #77

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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##:

$$x= -3, y = 1, z = 2\ \ \text{with} \ f(x) = f(y) = f(z) = -6$$

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##:

$$x= -3, y = 1, z = 2\ \ \text{with} \ f(x) = f(y) = f(z) = -6$$

Last edited:

- #78

fresh_42

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https://www.physicsforums.com/threads/math-challenge-march-2020-part-ii.985311/

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