I wrote
n2+2008=(n+2008)2−2⋅2008n−20082+2008=(n+2008)2−2⋅2008(n+2008)+20082+2008=(n+2008)2−2⋅2008(n+2008)+2008⋅2009
to deduce that n+2008 divides n2+2008 if and only if n+2008 divides 2008⋅2009.
Similarly, I found that n+2009 divides n2+2009 if and only if n+2009 divides 2009⋅2010.
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,m∈Z
Case 1: k=1 or m=1
(n+2008)=n2+2008 or (n+2009)=n2+2009
in any case leads to n2−n=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+2009−1)k=(n+2009)k−k≥(n+2009)(m+1)−k=(n+2009)m+(n+2009−k)≥n2+2009 also impossible, the last inequality holds because k<n+2009, suppose not
k≥n+2009>n+2008⇒n2+2008=(n+2008)k>(n+2008)2>n2+2008 a contradiction
Case 4: k=m>1
(n+2008)k+1=(n+2009)k⟹k=1⟹n=1
So the solutions are n=1 and n=0
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