Hello,
Prove that if (a/n) is irreducible then (a/n)^2 will also be irreduicble for a =/= 1, 0, n and n =/= 1, 0, a.
Thanks.
I have a direct proof of this.
Suppose a/n is irreducible. Therefore a and n share no common factor.
Factors of a:
Factors of b:
Now note that the quotient of any two factors of b and a
will
not be an integer.
*sub-proof:
*Suppose there does exist an i and j such that
, then this would imply that
. If some x is a factor of y, and y is a factor of z, then x is a factor of z. (Y=kx, z=ly = (lk)x). Therefore it follows that since b_j is a factor of a_i, then b_j is a factor of a, which is a contradiction since A and B share no factors.
— back to the proof
If there is no integer quotient
, then the square of this will also not be an integer,
is not an integer. And more generally, no product of quotients is an integer. But every factor of a^2 is either a factor of a or a product of 2 factors. So if there is no integer result to
, then there is no integer result to
, therefore
is irreducible**.
** That last statement is not true generally but is true in relation to factors. I could prove that but it is a bit complex.