Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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
2  
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
add comment

closed as not a real question by Hagen von Eitzen, Micah, Dennis Gulko, Davide Giraudo, Chris Eagle Mar 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.

2 Answers

up vote 2 down vote accepted

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 $1$.

share|improve this answer
add comment

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.

share|improve this answer
    
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
    
OK! And your answer, answers both possible readings. –  P.. Mar 8 '13 at 22:23
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.