Normalising Beal's conjecture Beal's conjecture
Can it be shown that

If
  $$  \sum_{q=0}^{u}(n+qd)^{m_{q}} =a^b $$
,where $n,u,d,a,m_q$ and $b$ are positive integers with $m_q,b> 3$, then $n,n+d,n+2d,...,n+ud$ and $a$ have a common prime factor.

Example


*

*$(n,u,d,a)=(98,2,98,98)$ and $(m_0,m_1,m_2,b)=(4,4,4,5)$

Related post 
Can it be shown, $n^4+(n+d)^4+(n+2d)^4\ne z^4$?
Extending Fermat's Last Theorem
Is there any solution for $n^x+(n+d)^y+(n+2d)^z=a^b$
 A: No. For example, $3^3 + 4^3 + 5^3 = 6^3$ and $\gcd(3,4,5,6) = 1$. You can find other examples from the identity
$$ 1^3 + 2^3 + \cdots + n^{3} = \left(\frac{n(n+1)}{2}\right)^{2}. $$
Since there are infinitely many triangular numbers that are also squares, there are infinitely many integers $n$ for which the right hand side of the above equation is a fourth power. (For example, $1^{3} + 2^{3} + 3^{3} + \cdots + 49^{3} = 35^{4}$.)
EDIT: The OP requested that all the exponents are at least $4$. Here's an example in that case:
$$ 1^4 + 2^4 + 3^4 + 4^4 + 5^4 + 6^4 + 7^4 + 8^4 + 9^4 + 10^5 + 11^5 + 12^5 + 13^4 + 14^4 + 15^5 + 16^4 + 17^4 = 35^4.$$
In general, one could fix an integer $N$ and consider expressions of the form $\sum_{q=1}^{N} q^{m_{q}}$ where $m_{q} \in \{k, k+1 \}$. There are $2^{N}$ such expressions, and all of these are between about $\frac{N^{k+1}}{k+1}$ and $\frac{N^{k+2}}{k+2}$. When $N$ is large enough (in terms of $k$), $2^{N}$ is much larger than $\frac{N^{k+2}}{k+2} - \frac{N^{k+1}}{k+1}$ and the chances of getting at least one $k$th power in that range are quite high.
