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Eclair_de_XII

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## Homework Statement

"If no prime number ##p## divides a hypothetical solution ##(x,y,z)∈ℕ×ℕ×ℕ## to the equation ##x^3+y^3=z^3##, prove that exactly one of x, y and z is even."

## Homework Equations

Given:

~##∃p:(\frac{x}{p},\frac{y}{p},\frac{z}{p})∈ℕ×ℕ×ℕ## such that ##x^3+y^3=z^3##.

In other words, ##(x,y,z)≠(ap,bp,cp)## for some ##a,b,c∈ℕ##

Prove: ##x##, ##y##, or ##z## can be written as a multiple of ##2##.

## The Attempt at a Solution

So here I assume that ##x, y, z## do not all have a common prime factor ##p## in common. I also assume that these variables are multiples of certain prime numbers or are prime numbers themselves. Taking these assumptions into mind, I set my variables: ##x=ap##, ##y=bp##, ##z=cp_u##. I assume that ##p_u≠p## is an arbitrary prime number. Now plugging in these values for the equation, I have...

##x^3+y^3=z^3##

##a^3p^3+b^3p^3=c^3p_u^3##

##p^3(a^3+b^3)=p_u^3(c^3)##

Now let ##a## be odd, and ##b## be odd. Then I can rewrite them as ##a=2m+1## and ##b=2n+1##. Focusing on the factor of ##p^3##, and assuming that it is not ##1##...

##(a)^3=(4m^2+4m+1)(2m+1)=(8m^3+8m^2+2m)+(4m^2+4m+1)=8m^3+12m^2+6m+1##

##(b)^3=(4n^2+4n+1)(2n+1)=(8n^3+8n^2+2n)+(4n^2+4n+1)=8n^3+12n^2+6n+1##

##a^3+b^3=8(m^3+n^3)+12(m^2+n^2)+6(m+n)+2=2[4(m^3+n^3)+6(m^2+n^2)+3(m+n)+1]##

Since ##[4(m^3+n^3)+6(m^2+n^2)+3(m+n)+1]## is an integer, then ##a^3+b^3## is even. Moreover, since ##p^3## only has itself and ##1## as a factor, it stands to say that either ##p_u^3## or ##c^3##--that is to say, either ##p_u## or ##c## are even. In turn, ##z## is an even integer, when ##x## and ##y## are both odd.

Now I'm a bit ambivalent about submitting this as a solution. Should I provide a case-by-case proof, or is this enough? Should I demonstrate that this does not work when ##x## or ##y## are odd; or perhaps should I make either ##x## or ##y## even, and the remaining of the two odd, with ##z## being odd? Or perhaps I should rework the problem so that only one (or perhaps none) of the three variables is a multiple of the prime number ##p##? Maybe if they were all distinct primes...? I don't know how many examples I should give for a case-by-case proof for this problem.

Additionally, I will need to prove that one of these terms is a multiple of ##3## with the same conditions stated above.

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