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| An Integer Root Problem (Posted on 2006-11-17) |
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Determine whether or not x²+7x-14(q²+1)=0
has any integer roots for integer q.
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Submitted by K Sengupta
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Rating: 5.0000 (1 votes)
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Solution:
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x^2 +7x - 14(q^2+1)= 0
or, x = ( -7 +/- sqrt(D))/2, where:
D = 49 + 56(q^2+1)
Now, for x to correspond to an integer, D must correspond to a perfect square. Since D is divisible by 7, it follows that:
49 + 56(q^2+1) = (7p)^2, for some integer p
or, 8*q^2 + 15 = 7* p^2
or, q^2+ 1 = 0( Mod 7)
or, q^2 = 6 (Mod 7)
Now, we know that 6 does not correspond to a quadratic residue in the Mod 7 system.
Consequently, no integer roots exist for integer q.
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