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.
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 0^{3} + 1^{3} = 1^{3} is a solution, and that 0 is divisible by 7.
another solution is (-a)^{3} + a^{3} = 0^{3} (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):
0^{3} = 0 (mod 7)
1^{3} = 1 (mod 7)
2^{3} = 1 (mod 7)
3^{3} = 6 (mod 7)
4^{3} = 1 (mod 7)
5^{3} = 6 (mod 7)
6^{3} = 6 (mod 7).
since x^{3} + y^{3} = z^{3}, z^{3} = 0,1, or 6 (mod 7).
since if z^{3} = 0 (mod 7) there is nothing to prove, assume that z^{3} = 1 or 6 (mod 7). then we have just two cases:
z^{3} = 1 (mod 7).
checking the possibilities (mod 7) we have:
0+0 = 0
0+1 = 1 <--this works, so x^{3} = 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 z^{3} = 6 there is also just one possibility:
0+6 = 6
here, again x^{3} = 0 (mod 7), so x = 0 (mod 7).