Let a, b and c be odd positive integers . Show that the quadratic equation ax2+bx+c=0 has no rational solution.

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Emily Green · Accepted Answer

Step 1We show that it either has zero or two rational solutions. Assume then contrary, that is&amp;nbsp;x1∈Q,x2∈R∖Q. Then&amp;nbsp;x1x2∈R∖Q, or&amp;nbsp;−ac∈R∖Q. Contradiction.Step 2Assume that is has two rational solutions. So it can be written as:&amp;nbsp;(x−n1m1)(x−n2m2)=0(m1x−n1)(m2x−n2)=0m1m2x2−(n1m2+n2m1)x+n1n2=0Now I will claim that we are done. Note that we can pick&amp;nbsp;ni,mi&amp;nbsp;such that&amp;nbsp;gcd(ni,mi)=1. We need the coefficients of x to have the same parity. If&amp;nbsp;m1&amp;nbsp;is even then&amp;nbsp;n2&amp;nbsp;must be even which gives&amp;nbsp;n1m2+n2m1&amp;nbsp;odd. The symmetric argument applies when&amp;nbsp;m2&amp;nbsp;is even. Finally, if both m are odd then n are odd but now&amp;nbsp;(n1m2+n2m1)&amp;nbsp;is even so we reach the required contradiction.

Let a, b and c be odd positive

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