Proof of ” perfect squares <=> a number which has an odd number of factors”

Proof of ” perfect squares \[\Rightarrow\] a number which has an odd number of factors” Let a perfect square n=\[k^2\], k is a natural number. Let \[k=p_0^{r_0}p_1^{r_1}\]…\[p_m^{r_m}\] where \[p_0,p_1,...,p_n\] are primes and \[p_i<p_i+1\] Then \[n=k^2\] \[=(p_0^{r_0}p_1^{r_1}\]…\[p_m^{r_m})^2\] \[=p_0^{2{r_0}}p_1^{2{r_1}}\]…\[p_m^{2{r_m}}\] Let d be a factor of n, then \[d==p_0^{2{x_0}}p_1^{2{x_1}}\]…\[p_m^{2{x_m}}\] where \[ 0<=x_i<=2{r_i}\], \[ 0<=i<=m\] We need to count the number of d, which is equal to the combination of factors of k and the power of factors of k. Let the total number be S, since \[ 0<=x_i<=2{r_i}\], we have \[S=(2{r_0}+1)(2{r_1}+1)...(2{r_m}+1)\] (since number of \[x_i\] is \[ 2{r_i}-0+1=2{r_i}+1\]) =oddodd…*odd =odd \[\Rightarrow \] ” perfect squares \[\Rightarrow\] a number which has an odd number of factors” Proof of ” perfect squares \[\Leftarrow\] a number which has an odd number of factors” Let \[n=p_0^{b_0}p_1^{b_1}\]…\[p_m^{b_m}\] where \[p_0,p_1,...,p_n\] are primes and \[p_i<p_i+1\] Let d be a factor of n, then \[d==p_0^{x_0}p_1^{x_1}\]…\[p_m^{x_m}\] where \[ 0<=x_i<=b_i\], \[ 0<=i<=m\] The number of d, which is equal to the combination of factors of k and the power of factors of n Let the total number be S \[S=(b_0+1)(b_1+1)...(b_m+1)\]=an odd number \[\Rightarrow\] \[ {b_i+1}\] is an odd number \[\Rightarrow \] \[b_i=2{r_i}\] where\[r_i\] is a natural number \[\Rightarrow \] \[n=p_0^{2{r_0}}p_1^{2{r_1}}\]…\[p_m^{2{r_m}}\] \[\Rightarrow \] \[n=(p_0^{r_0}p_1^{r_1}\]…\[p_m^{r_m})^2\] \[\Rightarrow \] n is a perfect square number.