no. of integeral solutions of x 2 +y 2 =2008

no. of integeral solutions of x2+y2=2008


1 Answers

mycroft holmes
272 Points
14 years ago

x2+y2 = 8X251 and 251 is a prime of the form 4k+3.


There is a result that states that if x2+y2 is divisible by a prime p of the form 4k+3, then both x and y are divisible by p.


But that means LHS is divisible by 2512 whereas RHS is divisible at most by 251 which is a contradiction.


Hence no integral solutions exist

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