Prove that for any integer $k>1$ and any positive integer $n$, there exist $n$ consecutive odd integers whose sum is $n^k$ Found these problems in a problem book and got stuck. The book doesn't have solutions to I've come here for help.
(1) Prove that for any integer $k>1$ and any positive integer $n$, there exist $n$ consecutive odd integers whose sum is $n^k$.
(2) Let $n$ be a positive integer and $m$ any integer of the same parity as $n$. Prove that there exist $n$ consecutive odd integers whose sum is $mn$.
 A: $(1)$  $$\sum_{r=0}^{n-1}(2a+2r-1)=\frac n2\left[2a-1+2a+2(n-1)-1\right]=n(2a+n-2)$$
$$\implies 2a+n-2=n^{k-1}\iff 2(a-1)=n(n^{k-2}-1)$$ which is even  for $k\ge2$ as then $n^{k-2}-1$ will be divisible by $n-1$
$(2)$ We need $n(2a+n-2)=mn\iff 2a+n-2=m\iff m-n=2a-2$ which is even $\implies m,n$ have same parity
A: Consider question part (1)
Without loss of generality, let $r$ denote an integer, so that the $n>0$ consecutive odd integers are 
$(2r+1),(2r+3),...,(2r + 2(n-1))$
The sum of these integers is 
$S=\sum_{g=0}^{n-1}(2r+1+2g)$  
This equals
$S=2rn+\sum_{g=0}^{n-1}(1+2g)=2rn+S_{odd}$
The second (summation) term is a sum of odd numbers $1,3,..,(2n-1)$, which evaluates to $n^2$, because
$S_{odd}=\sum_{g=0}^{n-1}(1+2g)=n+\frac{2n(n-1)}{2}=n^2$
Thus the sum of the $n$ consecutive numbers is 
$S = 2rn+n^2=n(2r+n)$
To show that $S=n^k$ for integer $k>1$ it suffices to show that we can find integer $r$ such that
$2r+n=n^{k-1}$
Rearranging we have
$\large r=\frac{n^{k-1}-n}{2}=\frac{n(n^{k-2}-1)}{2}$
The numerator of the RHS is always an even number for $k>1$, because if $n$ is even then $n^{k-2}$ is even as well, so that $n^{k-2}-1$ is odd, and the product of an even and odd number is even. Thus the RHS evaluates to an integer.
Conversely, if $n$ is odd, so is $n^{k-2}$, and $n^{k-2}-1$ is even, leading to the product of an even and odd number, which is even. Thus the RHS evaluates to an integer, so long as $k>1$.
Thus there exists integer $r$ such that $(2r+n)=n^{k-1}$, so that the sum of the $n$ consecutive numbers $S$ is $n^k$.
As for question part (2), recall that the sum of the $n$ consecutive integers (for intger $r,n$) is given by
$S=n(2r+n)$
Let us denote $m=(2r+n)$, then $m$ will be the same parity as $n$ as it equals $n$ plus an even number $2r$, so that
$S=nm$
A: I'm not sure why everyone is taking such long routes.
a. Special case of part $b$ since $n^{k-1}$ has the same parity as $n.$
b. If $m,n$ are odd, consider $m$ and $m \pm 2i, 1 \le i \le (n-1)/2.$ If $m,n$ are even, consider $m \pm (2i-1), 1 \le i \le n/2.$
