Determine if there is any triple of nonnegative integers, not necessarily different, (a,b,c) such that: a3+b3+c3=2016
Problem
Source: Mathematics Regional Olympiad of Mexico Northeast 2016 P1
Tags: number theory, Diophantine equation, diophantine
12.09.2022 22:05
WLOG a≥b≥c (symmetric equation). It is easy to see that a≥9 while c≤8. we also know that x^3 \equiv 0,1,8 \pmod 9 and as a^3 + b^3 + c^3 \equiv 0 \pmod 9, we get that either they all are divisible by 3, or one of them is and the other ones are \equiv 1,8 \pmod 9. Notice that if they all are divisible by 3, then 27|a^3 + b^3 + c^3 but 27\not \mid 2016 so not possible. Therefore only one of them is divisible by 3. We also notice that 13^3 > 2016 so a \leq 12. Therefore, a=9,10,11,12. If a=9, then b =9 because if b\leq 8, then 729 + 512 + c^3 \geq a^3 + b^3 + c^3 = 2016 \iff c^3 \geq 800 approx which isn't possible as c\leq b. So a=9 implies b=9 but we already discussed that this case isn't possible. Now a = 12 gives 2016 - 1728 = 288 = b^3 + c^3 which implies 6 \leq b < 7 but a=12,b=6 isn't possible as they both are divisible by 3. a=11 gives 685 = b^3 + c^3 implying 7 \leq b \leq 8. This also means c=3,6 as atleast one of them has to be divisible by 3 and clearly this doesn't give any solution. Now a=10 gives 1016 = b^3 + c^3 meaning 9 \leq b \leq 10. b=10 clearly not possible so b=9 meaning c=287 which isn't possible. Therefore no solution.
12.09.2022 23:39
Taking mod 7, we have a^3+b^3+c^3\equiv 0\pmod{7}. Now, examine the third power. All the possible third power modulo 7 are 0, 1, 1, -1, 1, -1, -1. Thus, the only way is to have 1, -1, 0 as a combination. With this constraint, plug in one of them to be the exponent of 0\pmod{7}, which can be either 0 or 7. Now, split into casework gives b^3+c^3=2016 or b^3+c^3=1673. Now if b^3+c^3=2016, then we have no solutions by bounding since b^3+c^3=(b+c)((b+c)^2-3bc) which we can let b+c=m, bc=n. This gives (m)(m^2-3n)=2016 and both m, n are integers. Try factors of m yields no solutions. Furthermore, if b^3+c^3=(b+c)(b^2+c^2-bc)=1673, then since both are integers one of them must be 7 and the other be 239, which is a prime. This clearly yields no solutions. Thus, there are no such triple. q.e.d.
13.09.2022 07:49
In the above solution, Would anyone enlighten me how to show 3a^3 \geq a^3 +b^3 +c^3?
13.09.2022 09:10
Vulch wrote: In the above solution, Would anyone enlighten me how to show 3a^3 \geq a^3 +b^3 +c^3? Because we wlog a\geq b\geq c\implies a^3\geq b^3\geq c^3.
14.09.2022 06:59
Vulch wrote: In the above solution, Would anyone enlighten me how to show 3a^3 \geq a^3 +b^3 +c^3? Bounding.