# The Lost Proof of Fermat

5^1 = 1*0 + 5

5^2 = 2*10 + 5

5^3 = 3*40 + 5

5^p = p*a + 5

x^p = p*a + x

x^p = p*a + x

y^p = p*b + y

z^p = p*c + z

...x^p + y^p = z^p

p*a + x + p*b + y = p*c + z

p*[a + b - c] = z - [x + y]

p = [z - (x + y)]/[a + b - c]

http://www.maa.org/mathland/mathtrek_12_8.html [Broken]

Astonishingly, a proof of the ABC conjecture would provide a way of establishing Fermat's last theorem in less than a page of mathematical reasoning. Indeed, many famous conjectures and theorems in number theory would follow immediately from the ABC conjecture, sometimes in just a few lines.

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how does this prove that x,y and z can't be integers when p>2? Also a,b and as far as I know can be anything you want.

FulhamFan3 said:
how does this prove that x,y and z can't be integers when p>2? Also a,b and as far as I know can be anything you want.

p, x, y, a,b,c, are always integers > 0, when p is a prime number. When p is 1, a = 0, b = 0, and c = 0.

z and c cannot be a +integer for p > 2.

[x^p - x]/p = a

[y^p - y]/p = b

[z^p - z]/p = c

for example:

[3^2 - 3]/2 = 3

[3^3 - 3]/3 = 8

[3^5 - 3]/5 = 48

[+integer^p - +integer]/p = another +integer.

It works for all prime numbers.

3^2 + 4^2 = 5^2

2*3 + 3 = 3^2

2*6 + 4 = 4^2

2*3 + 4*3 + 3 + 4 = 6*3 + 7 = 5^2

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Russel Rierson,

I don't know why this is here and no in the number theory section of the math board.

But as long as it's here, you lost me on your ifrst line: 5^1 = 1*0 + 5

Russell E. Rierson said:
p, x, y, a,b,c, are always integers > 0, when p is a prime number. When p is 1, a = 0, b = 0, and c = 0.

z and c cannot be a +integer for p > 2.

z and c can be integers when p>2. you also have to prove that statement you can't just say it.

Also fermat's theorem as far as I know isn't just for primes, it's for all numbers. Unless this is some other theorem by fermat. In which case you need to state it so we know what your talking about.

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FulhamFan3 said:
z and c can be integers when p>2. you also have to prove that statement you can't just say it.

Also fermat's theorem as far as I know isn't just for primes, it's for all numbers. Unless this is some other theorem by fermat. In which case you need to state it so we know what your talking about.

If p = 2 :

x^2 = 2*[1 + 2 + 3+...+ x-1] + x

2^2 = 2* + 2

3^2 = 2*[1 + 2] + 3

4^2 = 2*[1 + 2 + 3] + 4

5^2 = 2*[1 + 2 + 3 + 4] + 5

etc...

This congruence does not hold for p > 2

5^2 = 2*[1 + 2 + 3 + 4] + 5

5^2 = 2*[1 + 2] + 2*[3 + 4] + 5

5^2 = 2*[1 + 2] + 2*[3 + (4 - 1)] + 5 + 2*1

5^2 = 2*[1 + 2] + 2*[3 + 3] + 7

5^2 = 2*[1 + 2] + 3 + 2*[ 1 + 2 + 3] + 4

5^2 = 3^2 + 4^2