## ritvik1512 2 years ago Peter has recently been fascinated by numbers that can be expressed in the form f(a,b,c)=a(b−c)^3+b(c−a)^3+c(a−b)^3 for some integers a,b and c. He thinks that we should use this system to do arithmetic in future. How many positive integers N≤1000 can be expressed in the form of f(a,b,c)?

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1. goformit100

f(a,b,c)=a(b−c)^3+b(c−a)^3+c(a−b)^3

2. mathslover

Oh! Peter should try it first.

3. PeterPan

This is a lie.

4. PeterPan

Unless it's Spider-man or Peter Griffin :D

5. goformit100

Oh! Peter should definitely try it first.

6. mathslover

@ritvik1512 I meant to say that, can you show us what you tried so that we can get to your mistake and problem and help accordingly.

7. goformit100

This is Truly LIE.

8. goformit100

9. shubhamsrg

@shubhamsrg

10. electrokid

does anybody think this might give only the even numbers?

11. joemath314159

If you manipulate the equation a bit (by expanding the cubes, and simplifying/factoring), you can show that:\[f(a,b,c)=ab(b+a)(b-a)+bc(c+b)(c-b)+ac(a-c)(a+c)\]From here is is easier to show (using a modular arithmetic argument) that no matter what integers you pick, this will always be divisible by both 2 and 3 (so divisible by 6). My guess is that you can only get the multiples of 6, and nothing else.

12. joemath314159

In fact, note that for any integer a:\[f(a,a+1,a+2)=a(-1)^3 +(a+1)(2)^3 +(a+2)(-1)^3 \]\[-a+8a+8-a-2=6a+6=6(a+1)\]So you can get any multiple of 6 you want.