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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?

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  • $\begingroup$ There can't be one $k$ which divides $x$ and $x+1$, other than $k=1$. So it has to be a different square for each element. $\endgroup$ – Thomas Andrews Mar 8 '17 at 19:48
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    $\begingroup$ Do you know the Chinese remainder theorem? Are you allowed to find any solution, or does it have to be the smallest? $\endgroup$ – Thomas Andrews Mar 8 '17 at 19:53
  • $\begingroup$ I can find any solution $\endgroup$ – Temirzhan Mar 8 '17 at 19:54
  • $\begingroup$ @ThomasAndrews Except that two of $x, \ldots, x+5$ may be divisible by $4$. $\endgroup$ – Robert Israel Mar 8 '17 at 19:58
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    $\begingroup$ You say 5 consecutive integers, but $x,x+1,\ldots,x+5$, is 6 consecutive integers. $\endgroup$ – B. Goddard Mar 8 '17 at 20:00
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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 $49$.

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  • $\begingroup$ Very mechanical, though, and since this implies an answer less than $4\cdot9\cdot25\cdot49\cdot121 = 5336100$, all of the calculations can be done straightforwardly using standard 32-bit integer arithmetic. $\endgroup$ – Steven Stadnicki Mar 8 '17 at 20:03
  • $\begingroup$ And it's even easier if the OP really mean $5$ consecutive, rather than six. $\endgroup$ – Thomas Andrews Mar 8 '17 at 20:08
  • $\begingroup$ Thomas what about a numver thats congruent to -4? you have 0, -1, -2, -3, -5 but you skip -4 $\endgroup$ – Temirzhan Mar 8 '17 at 20:13
  • $\begingroup$ If $x\equiv 0\pmod{4}$ then $x\equiv -4\pmod{4}$. @user203042 $\endgroup$ – Thomas Andrews Mar 8 '17 at 20:14
  • $\begingroup$ They don't have to be "in order." $\endgroup$ – Doug M Mar 8 '17 at 20:17
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The formulation you give ($x,x+1,\ldots,x+5$) actually has $6$ consecutive values. The values for such $x$ below $100000$:

$$22020, 24647, 30923, 47672, 55447, 57120, 73447, 74848, 96675$$

Finding $5$ consecutive values is easier; the smallest such sequence is $844,845,846,847,848$, which are multiples of $4,169,9,121,16$ respectively.

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  • $\begingroup$ Interesting. $844 \neq 29348(mod4⋅9⋅24⋅49)$ [sic] as stated in @ThomasAndrews post. $\endgroup$ – Χpẘ Mar 8 '17 at 20:57
  • $\begingroup$ This is the result of a straight search, so as you can see one of the squares involved is $13^2=169$ and the other factor squares are in a different order. $\endgroup$ – Joffan Mar 8 '17 at 20:57

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