Lesson Quadratic polynomial with odd integer coefficients can not have a rational root

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<H2>Quadratic polynomial with odd integer coefficients can not have a rational root</H2>

<H3>Problem 1</H3>Given odd integers &nbsp;a, &nbsp;b, &nbsp;c, &nbsp;prove that the equation &nbsp;{{{ax^2+bx+c}}} = {{{0}}} &nbsp;cannot have a solution x which is a rational number.

<B>Solution</B>

<pre>
Assume the equation {{{ax^2+bx+c}}} = {{{0}}} with odd integer coefficients a, b an c has the solution, 
which is a rational fraction {{{p/q}}} with integer p and q.

We can assume that all the common divisors of p and q are just canceled in the fraction {{{p/q}}}, 
so that p and q are relatively primes integer numbers. In particular, p and q are not both multiples of 2 simultaneously.

Then substitute the fraction {{{p/q}}} into the equation.

You will get {{{a*(p/q)^2 + b*(p/q) + c}}} = {{{0}}}.

Multiply both sides by {{{q^2}}} to rid off the denominators. You will get

{{{a*p^2 + b*pq + c*q^2}}} = {{{0}}}.   (1)

Now, if p is odd, then q can not be multiple of 2, otherwise you easily get a contradiction due to equation (1).

Similarly, if q is odd, then p can not be multiple of 2, otherwise you easily get a contradiction due to equation (1).

Thus both p and q must be odd.

Then the equation (1) has three odd addends that sum up to zero, which is impossible.

This contradiction completes the proof.
</pre>

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