Thread: Showing something is divisible

Hey there I need some help with this problem.

If x^3 + y^3 = z^3, show that one of x,y,z is divisible by 7

Just wondering if someone can show me how to do this problem, so I can understand how to do it thanks.

Consider this equality modulo 7.

The question assumes that x,y and z are integers. So, by Fermat's Last Theorum(FLM) there are no solutions to the equation, ie no solution divisible by 7.



"FLM:There exists no solutions to the equation x^n +y ^n = z^n for n>2, x,y,z are integers and xyz≠0

It is a strange question, or have I missed something?

note that 0³ + 1³ = 1³ is a solution, and that 0 is divisible by 7.

another solution is (-a)³ + a³ = 0³ (again, 0 is divisible by 7).

this is not a special case of FLT.

emakarov's suggestion is pertinent, here, let's look at the cubes (mod 7):

0³ = 0 (mod 7)
1³ = 1 (mod 7)
2³ = 1 (mod 7)
3³ = 6 (mod 7)
4³ = 1 (mod 7)
5³ = 6 (mod 7)
6³ = 6 (mod 7).

since x³ + y³ = z³, z³ = 0,1, or 6 (mod 7).

since if z³ = 0 (mod 7) there is nothing to prove, assume that z³ = 1 or 6 (mod 7). then we have just two cases:

z³ = 1 (mod 7).

checking the possibilities (mod 7) we have:

0+0 = 0
0+1 = 1 <--this works, so x³ = 0 (mod 7), thus x = 0 (mod 7) (since the only thing (mod 7) when cubed that gives 0, is 0).
0+6 = 6
1+1 = 2
1+6 = 0
6+6 = 5

(this is all we need to consider since addition mod 7 is commutative).

for z³ = 6 there is also just one possibility:

0+6 = 6

here, again x³ = 0 (mod 7), so x = 0 (mod 7).

I see now thanks a lot for your help