Question

If`2m^2=p^2` Prove that `2`is a factor of `p`?

Answer 1

`"See explanation"`

Explanation:

`"Suppose p is odd so that 2 is not a factor of p."`
`"Then p can be written as 2n+1."`
`=> p^2 = (2n+1)^2 = 4n^2 + 4n + 1`
`"Now "(4 n^2 + 4n + 1)" mod 2 = 1,"`
`"so "p^2" is odd."`
`p^2 = 2 m^2 " is impossible as such as "2 m^2" is even."`
`"Hence our assumption that p is odd is false, so p must be even."`

`"One can also work through the prime factorization that is"`
`"unique :"`
`p^2 " contains 2 in its prime factorization."`
`"Hence also "p" contains 2 in its prime factorization as a square"`
`"of a number has the same prime factorization but with the"`
`"exponents doubled."`