# consecutive number such that divide some square number bigger one [closed]

how to prove $\forall k\in \Bbb Z$ ,$k\ge 1$ ,there exists $k$ consecutive number such that divide some square number bigger one.

it's seem we must use Chinese remainder theorem .but how?

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## closed as not a real question by Hagen von Eitzen, Micah, Dennis Gulko, Davide Giraudo, Chris EagleMar 9 '13 at 16:34

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

I can't parse the statement you are trying to prove. What does "some square number bigger one" mean? –  Asaf Karagila Mar 8 '13 at 22:02
Agreed. Can you write out an example when, say, $k=3$, to make clear what you mean? –  Thomas Andrews Mar 8 '13 at 22:03
Maybe $\forall k\ge 1\exists n\forall 0\le i<k\exists m>1\colon m^2|(n+i)$? –  Hagen von Eitzen Mar 8 '13 at 22:06
@HagenvonEitzen: The question is unclear. Your interpretation may well be what was meant, but when I read the question, I read it as Ross Millikan did: $\forall k\ge1,\exists n\gt1,j:\forall 1\le i\le k,(j+i)\mid n^2$ –  robjohn Mar 9 '13 at 0:01

Let $p_1,\dots,p_k$ be distinct primes.

Look at the system of congruences $x\equiv 0\pmod{p_1^2}$, $x+1\equiv 0\pmod{p_2^2}$, and so on up to $x+k-1\equiv 0\pmod{p_k^2}$.

By the Chinese Remainder Theorem, this system of congruences has a solution $x$. Note that for $i=0$ to $k-1$, the number $x+i$ is divisible by $p_i^2$, so is divisible by a square greater than $1$.

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How about just saying that $(k!)^2$ is divisible by all of $\{1^2,2^2,3^2 \ldots k^2\}$? Maybe I didn't understand the question, but if so, please explain what it is.

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I agree with you, but maybe you mean $\{1,2,3,\ldots,k\}$? –  P.. Mar 8 '13 at 22:13
@P..: That was my original thought. We are still trying to understand the question. André Nicolas and I have proposed answers to two possible readings of the question. Maybe OP will tell us which is correct or give a third possibility. –  Ross Millikan Mar 8 '13 at 22:20