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]

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