anonymous
  • anonymous
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)?
Precalculus
schrodinger
  • schrodinger
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goformit100
  • goformit100
f(a,b,c)=a(b−c)^3+b(c−a)^3+c(a−b)^3
mathslover
  • mathslover
Oh! Peter should try it first.
anonymous
  • anonymous
This is a lie.

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anonymous
  • anonymous
Unless it's Spider-man or Peter Griffin :D
goformit100
  • goformit100
Oh! Peter should definitely try it first.
mathslover
  • 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.
goformit100
  • goformit100
This is Truly LIE.
goformit100
  • goformit100
@nader1
shubhamsrg
  • shubhamsrg
@shubhamsrg
anonymous
  • anonymous
does anybody think this might give only the even numbers?
anonymous
  • anonymous
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
  • anonymous
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.

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