Consider there are five consecutive numbers $a,b,c,d$ and $e$ such that $a \lt b \lt c \lt d\lt e$.Given that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube, find the least value of $c$.
I just want to know how to approach these kind of problems.
Thanx for any help. :)
Restating @Lord Shark's hint, we have that $3c$ must be a perfect square and $5c$ a perfect cube.
The smallest such number $c$ would obviously have prime factors of the form $3^x5^y$.
Now, $3^{x+1}5^y$ is a perfect square $\implies$ $x + 1$ and $y$ are even.
And, $3^x5^{y+1}$ is a perfect cube $\implies$ $x$ and $y + 1$ are divisible by $3$.
Can you solve for the smallest such $x$ and $y$?