Consecutive integers

$a<b<c<d<e$ are five consecutive integers,so set $a=c-2,b=c-1,d=c+1,e=c+2$ .Therefore we have for some integers $m,n$ : $\begin{cases} b+c+d=m^2\implies 3c=m^2\qquad (1)\\ a+b+c+d+e=n^3\implies 5c=n^3\qquad (2) \end{cases}$ From $(1)$ we get that $c=3k^2$ for some $k$ .

Putting this in $(2)$ yields $15k^2=n^3$ .Hence $k$ must be of the form $k=15p^3$ for some $p$ .

Hence $c=3k^2=3(15p^3)^2=675p^6$ .The minimum value of $c$ will occur when $p=1$ .Hence the samllest possible value of $c$ is $c=675p^6=\boxed{675}$

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For some who might be a bit confused about why I put $c=3k^2$ or $k=15p^3$ .The reason is as follows.I'll explain why I put $c=3k^2$ .The expression for $k$ follows similarly.

From $(1)$ we get $c=\dfrac{m^2}{3}$ .Now,we are given that $c$ is a positive integer.But that means that $c=\dfrac{m^2}{3}$ is a positive integer.Now what does that say about $m$ ? Think about it.Our main problem is that 3 in the denominator.If $c$ is an integer then upon simplifying $\dfrac{m^2}{3}$ ,that 3 in the denominator must get cancelled out.Now that will only happen if $m^2$ has a factor of 3 in its factorization,which means that $m$ is a multiple of 3.Therefore,setting $m=3k$ ,we get $c=\dfrac{m^2}{3}=\dfrac{(3k)^2}{3}=\dfrac{9k^2}{3}=3k^2$ ,as desired.

Let $a < b < c < d < e$ be five consecutive positive integers such that $b = a+1, c= a+2, d = a+3, e = a+4$ . If we require the smallest value of $c$ such that $b + c + d = 3a + 6 = \alpha^2$ and $a +b +c + d + e = 5a + 10 = \beta^3$ , then we have:

$a + 2 = c = \frac{\alpha^2}{3} = \frac{\beta^3}{5} \Rightarrow 5\alpha^{2} = 3\beta^{3}.$ (i)

We require the prime factorizations $\alpha = 3^{p_{1}}5^{p_{2}}$ and $\beta = 3^{p_{3}}5^{p_{4}}$ , which after substituting these into (i) yields:

$3^{2p_{1}}5^{2p_{2} + 1} = 3^{3p_{3} + 1}5^{3p_{4}} \Rightarrow 2p_{1} = 3p_{3} + 1, 2p_{2} + 1 = 3p_{4}$

which the smallest permissible positive integral values are $p_{1} = 2, p_{2} = p_{3} = p_{4} = 1.$ The smallest possible values of $\alpha$ and $\beta$ are $\alpha = 3^{2}5^{1} = 45, \beta = 3^{1}5^{1} = 15$ , and the smallest possible value for $c$ computes to:

$c = \frac{\alpha^2}{3} = \frac{\beta^3}{5};$

or $c = \frac{45^2}{3} = \frac{15^3}{5} = \boxed{675}.$