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)?

Question

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)?

goformit100 · Answer

f(a,b,c)=a(b−c)^3+b(c−a)^3+c(a−b)^3

mathslover · Answer

Oh! Peter should try it first.

anonymous · Answer

This is a lie.

anonymous · Answer

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

goformit100 · Answer

Oh! Peter should definitely try it first.

mathslover · Answer

@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.

goformit100 · Answer

This is Truly LIE.

goformit100 · Answer

@nader1

shubhamsrg · Answer

@shubhamsrg

anonymous · Answer

does anybody think this might give only the even numbers?

anonymous · Answer

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.

anonymous · Answer

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.