Question

Let M ∈ N >1 . Let M be a non-perfect square. Or in other words M /∈ {4, 9, 16, · · · } (a) Prove that there exists a prime p, whose exponent in the prime factorization of M is odd. [Hint: Prove by a method of contradiction]

Answer 1

Explanation:

By the fundamental theorem of arithmetic given
`n\in \mathbb{N}_(\ge2)` there exists primes
`p_(i)` and integers
`\alpha_i` such that
`n` can be written as
`n=p_1^(\alpha_1)p_2^(\alpha_2)p_3^(\alpha_3)\cdots p_t^(\alpha_t)`, now suppose that
`n` is a non-perfect square, we are going to prove that there exists
`\alpha_i` such that is odd. By contradiction, suppose that
`\alpha_i` is even for all
`i`, then writing
`\alpha_i=2\beta_i` for all
`i`, we can write
`n=p_1^(\alpha_1)p_2^(\alpha_2)p_3^(\alpha_3)\cdots p_t^(\alpha_t)=  p_1^(2\beta_1)p_2^(2\beta_2)p_3^(2\beta_3)\cdots p_t^(2\beta_t)=(p_1^(\beta_1)p_2^(\beta_2)p_3^(\beta_3)\cdots p_t^(\beta_t))^2`, thus we conclude that
`n` is perfect square, a contradiction, and then we conclude that there exists
`\alpha_i` such that is odd.