Five Consecutive Integers divisible by a square greater than 1 I have the task of finding 5 consecutive 
integers of the 
form {x, x+1, x+2, x+3, x+4, x+5}
where each number in the sequence 
is divisible by a square  k greater than 1..
I tried to write a simple JAVA program
to find a sequence of that form by checking
if each number in the sequence = 0 mod k
But that can't be possible, how would I go about
finding these numbers?

Chinese remainder theorem:
$$\begin{align}x&\equiv 0\pmod{4}\\
x&\equiv -1\pmod{9}\\
x&\equiv -2\pmod{25}\\
x&\equiv -3\pmod{49}\\
x&\equiv -5\pmod{121}
\end{align}$$
That's gonna be ugly.
If you only need five, then you can ignore the last line and you get $$x\equiv 29348\pmod{4\cdot 9\cdot 24\cdot 49}.$$ Then $x,x+4$ are divisible by $4$, $x+1$ is divisible by $9$, $x+2$ is divisible by $25$, and $x+3$ is divisible by $49$. 
