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Question

Let p and q are distinct naturals such that 2005+p=q2 and 2005+q=p2.Find the value of 2014+pq

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Solution

(2005+p)-(2005+q)=q^2-p^2
p-q=(q-p)(p+q)
p+q=(-1)
p^2+q^2=p+q+4010=4009
(p+q)^2=p^2+q^2+2pq=1
4009+2pq=1
pq=(-2004)
2014+pq=2014-2004=10

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