Q27 from AMC 2012(Senior): Five consecutive integers that sum to a perfect square, and the three middle terms sum to a perfect cube. 
Five consecutive integers $p,q,r,s,t$,each less than $10000$, produce a sum which is a perfect square, while the sum of $q,r,s$ is a perfect cube.What is the value of $ \sqrt{p+q+r+s+t}$ ?

What I have tried so far:
Let $p=r-2$
$p+q+r+s+t =5r $
$5r=x^2 $
$q+r+s =3r =y^3 $
$x^2 -y^3 =2r $
So,the only perfect squares which are divisible by $5$ are the multiples of $5$: $25,100,225,400...$
I also observed that $100-25=75,225-100=125$,where $125-75=50$.Trying that for $225-100=125,400-225=175$ where $175-125=50$
Then,for the perfect cube which are divisible by 3 and must be less than the perfect squares.
$30^3,60^3,90^3,120^3....$
And here is were I got stuck at...
Is the concept I'm using correct?

Since $\frac{3}{5}x^2=y^3$ is a cube, the number of factors of 5 in $x^2$ must be even and one more than a multiple of 3. So choices are $4$ and $10$. Also the number of factors of 3 in $x^2$ must be even and one less than a multiple of $3$, so choices are $2$ and $8$. Now, $\frac{x^2}{5} \le 10000$, so $x$ must be  a  multiple of $3\cdot 5^2 = 75$. And in fact $x=75$ (so $r = \frac{75^2}{5}=1125$) works.
