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.

Printable View

- November 29th 2012, 09:40 PMgfbrdShowing 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. - November 30th 2012, 01:59 AMemakarovRe: Showing something is divisible
Consider this equality modulo 7.

- November 30th 2012, 02:10 AMTychoBraheRe: 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? - November 30th 2012, 06:10 AMDevenoRe: Showing something is divisible
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). - November 30th 2012, 02:04 PMgfbrdRe: Showing something is divisible
I see now thanks a lot for your help