383this question was already discussed, ans is (A)
Is there any shorter way to solve this ques.
Consider the equation x2+y2 = 2007. How many solutions (x; y) exist
such that x and y are positive integers?
(A) None
(B) Exactly two
(C) More than two but
nitely many
(D) In
fitely many.
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3 Answers
1057Reason:
x and y both can't be even or both odd(since sum of two even or two odd numbers is always even)
So even if at all a solution exists one has to be even and one has to be odd.
let x=2k
y=2r+1
x2+y2=4k2+(2r+1)2 which is 4(k2+r2+r)+1 or of the form 4p+1
since 2007 is not of the above form,no solutions.
p.s:here r,k,p all are integers...
