Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.,
Is
$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4$$ true $\forall $ $n \in \mathbb{N}$

Let the LHS be $A(n)$ and let the RHS be $2B(n)^4.$ 
Then $A(n+1)-A(n)=(n+1)^7+(n+1)^5=(n+1)^5(n^2+2n+2).$ 
$$\text {We have  }\quad 2B(n+1)^4-2B(n)^4=$$ $$(*)\quad =2(B(n+1)^2+B(n)^2)\cdot (B(n+1)+B(n))\cdot (B(n+1)-B(n)).$$  Since $B(n)=n(n+1)/2$ we have $$B(n+1)^2+B(n)^2=(n+1)^2((n+2)^2+n^2)/4=(n+1)^2(n^2+2n+2)/2$$ $$\text { and}\quad B(n+1)+B(n)=(n+1)((n+2)+n)/2=(n+1)^2$$ $$\text { and }\quad B(n+1)-B(n)=n+1.$$ From these we compute $(*)$ (4th line from the top) and find it is equal to $(n+1)^5(n^2+2n+2)$, which is $A(n+1)-A(n).$ (2nd line from the top.). 
So if $A(1)=B(1)$ then $A(n)=B(n)$ for all $n\in \Bbb N$ by induction on $n.$
Interesting identity. Another interesting one is $\sum_{x=1}^nx^3=(\sum_{x=1}^nx)^2=(n(n+1)/2)^2.$
