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This is what the theorem states:
"No three positive integers a, b, and c can satisfy the equation a^p+b^p=c^p, for any integer values of p >2"
My attempt (p = n+2 > 2):
a^{p}+b^{p}=c^{p}
a^{(n+2)}+b^{(n+2)}=c^{(n+2)}
(a^{n}*a^{2 })+(b^{n}*b^{2} )=(c^{n}*c^{2} )
(a^{n}*a^{2})/(a^{n}*b^{n}*c^{n} )+(b^{n}*b^{2})/(a^{n}*b^{n}*c^{n})=(c^{n}*c^{2})/(a^{n}*b^{n}*c^{n})
a^{2} (1/bc)^{n} + b^{2} (1/ac)^{n} = c^{2} (1/ab)^{n}
The equation is now similar to the Pythagorean sum, and in order for it to remain true we must have the following:
a^{2} (1) + b^{2} (1) = c^{2} (1)
(1/bc)^{n}=(1/ac)^{n}=(1/ab)^{n}=1
∴n=0 or a=b=c=1
If a=b=c=1, we will have the following:
1^{p}+1^{p}=1^{p}
2=1
Therefore, a=b=c=1 cannot be true, hence n = 0 is the remaining option which means:
p=n+2
p=0+2
p=2
∴p cannot be greater than 2
This is what the theorem states:
"No three positive integers a, b, and c can satisfy the equation a^p+b^p=c^p, for any integer values of p >2"
My attempt (p = n+2 > 2):
a^{p}+b^{p}=c^{p}
a^{(n+2)}+b^{(n+2)}=c^{(n+2)}
(a^{n}*a^{2 })+(b^{n}*b^{2} )=(c^{n}*c^{2} )
(a^{n}*a^{2})/(a^{n}*b^{n}*c^{n} )+(b^{n}*b^{2})/(a^{n}*b^{n}*c^{n})=(c^{n}*c^{2})/(a^{n}*b^{n}*c^{n})
a^{2} (1/bc)^{n} + b^{2} (1/ac)^{n} = c^{2} (1/ab)^{n}
The equation is now similar to the Pythagorean sum, and in order for it to remain true we must have the following:
a^{2} (1) + b^{2} (1) = c^{2} (1)
(1/bc)^{n}=(1/ac)^{n}=(1/ab)^{n}=1
∴n=0 or a=b=c=1
If a=b=c=1, we will have the following:
1^{p}+1^{p}=1^{p}
2=1
Therefore, a=b=c=1 cannot be true, hence n = 0 is the remaining option which means:
p=n+2
p=0+2
p=2
∴p cannot be greater than 2
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