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john !!
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#1
Report Thread starter 14 years ago
#1
find the sum of all p for which p^2 + 103 is a perfect square.
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J.F.N
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Report 14 years ago
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p^2 + 103 = r^2
--> r^2 - p^2 = 103
--> (r-p)(r+p)=103
103 is prime, so either
a) r-p=103 and r+p=1 --> r=52, p=-51
b) r-p=1 and r+p=103 --> r=52, p=51
So sum of all p=0
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