Monday, 24 June 2013

elementary number theory - Find all positive integers n such that n+2008 divides n2+2008 and n+2009 divides n2+2009



I wrote
n2+2008=(n+2008)222008n20082+2008=(n+2008)222008(n+2008)+20082+2008=(n+2008)222008(n+2008)+20082009
to deduce that n+2008 divides n2+2008 if and only if n+2008 divides 20082009.



Similarly, I found that n+2009 divides n2+2009 if and only if n+2009 divides 20092010.



I can't seem to get anywhere from here. I know that n=1 is an obvious solution, and I think it's the only one, although I can't prove/disprove this. I know that any two consecutive integers are coprime but I haven't been able to use this fact towards the solution.



Could anyone please give me any hints?




For the record, this is a question from Round 1 of the British Mathematical Olympiad.


Answer



Here another approach



Since (n+2008)|(n2+2008) and (n+2009)|(n2+2009) then



(n+2008)k=n2+2008 and (n+2009)m=n2+2009 for some k,mZ



Case 1: k=1 or m=1




(n+2008)=n2+2008 or (n+2009)=n2+2009



in any case leads to n2n=0 that is n=1 and n=0



Case 2: m>k>1



n2+2009=(n+2009)m=(n+2008)m+m>(n+2008)k+m>n2+2009 which is impossible



Case 3: k>m>1




n2+2008=(n+20091)k=(n+2009)kk(n+2009)(m+1)k=(n+2009)m+(n+2009k)n2+2009 also impossible, the last inequality holds because k<n+2009, suppose not



kn+2009>n+2008n2+2008=(n+2008)k>(n+2008)2>n2+2008 a contradiction



Case 4: k=m>1



(n+2008)k+1=(n+2009)kk=1n=1



So the solutions are n=1 and n=0


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