Question

Let a, b, c, d, e be consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube. Find the smallest possible value of c.

Answer 1

Since the middle term of an arithmetic progression with an odd number of terms is the average of the series,

we know $b+c+d=3c$ and $a+b+c+d+e=5c$.

Thus, $c$ must be in the form of $3\cdot x^2$ based upon the first part

and in the form of $5^2\cdot y^3$ based upon the second part, with $x$ and $y$ denoting an integers.

$c$ is minimized if it’s prime factorization contains only $3,5$,

and since there is a cubed term in $5^2\cdot y^3$, $3^3$ must be a factor of $c$.

$3^3{5}^2=675$