Determine whether or not x²+7x-14(q²+1)=0
has any integer roots for integer q.
If a and b are roots, then expand (x-a)(x-b) = 0 to find that (-a - b) = 7 and ab = -14(q^2 + 1).
Solve the first equation for b and substitute into the second equation to get a(a + 7) = 14(q^2 + 1).
So 7 factors a, and thus 7 factors (a + 7) and 7 factors (q^2 + 1).
But (q^2 + 1) <> 0 mod 7.
So there are no integer roots.
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Posted by xdog
on 2006-11-17 11:25:23 |
solution
