fermat

if a^x+b^x=c^x, x>=3 has no solutions for integers a,b,c, prove that c must be irrational if a and b are rational, or find a valid example where a,b and c are all rational. (e.g. 3^2+4^2=5^2 for x=2, but x>=3...)

the proof is actually surprisingly easy...either that, or i got it wrong 😕

a and b are rational. Therefore a can be written a1/a2, b as b1/b2 (a1,a2,b1,b2 all integers).

Assume that c is rational too. So c can be written as c1/c2 (c1 and c2 integers).

(a1/a2)^n + (b1/b2)^n = (c1/c2)^n

Multiply both sides by (a2*b2*c2)^n

This gives (a1*b2*c2)^n + (b1*a2*c2)^n = (c1*a2*b2)^n

Clearly a1*b2*c2, b1*a2*c2 and c1*a2*b2 are all integers and satisfy A^n + B^n = C^n, which we know has no integer solutions.

Therefore our assumption that c is rational must be false.

Originally posted by Fat Lady
a and b are rational. Therefore a can be written a1/a2, b as b1/b2 (a1,a2,b1,b2 all integers).

Assume that c is rational too. So c can be written as c1/c2 (c1 and c2 integers).

(a1/a2)^n + (b1/b2)^n = (c1/c2)^n

Multiply both sides by (a2*b2*c2)^n

This gives (a1*b2*c2)^n + (b1*a2*c2)^n = (c1*a2*b2)^n

Clearly a1*b2*c2, b1*a2*c2 and c1*a2*b2 are ...[text shortened]... we know has no integer solutions.

Therefore our assumption that c is rational must be false.

see-simple 🙂

EDIT: i'm bored now. therefore, next question,

can 1/2+1/3+...+1/n=A where A is an integer?

this one is -erm- slightly tougher...