# Showing something is divisible

• Nov 29th 2012, 10:40 PM
gfbrd
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.
• Nov 30th 2012, 02:59 AM
emakarov
Re: Showing something is divisible
Consider this equality modulo 7.
• Nov 30th 2012, 03:10 AM
TychoBrahe
Re: Showing something is divisible
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?
• Nov 30th 2012, 07:10 AM
Deveno
Re: Showing something is divisible
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).
• Nov 30th 2012, 03:04 PM
gfbrd
Re: Showing something is divisible
I see now thanks a lot for your help