SOLUTION: Given odd integers a, b, c, prove that the equation {{{ax^2+bx+c=0}}} cannot have a solution x which is a rational number.

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Question 1027170: Given odd integers a, b, c, prove that the equation cannot have a solution x which is a rational number.
Found 2 solutions by richard1234, ikleyn:
Answer by richard1234(7193)   (Show Source): You can put this solution on YOUR website!
Assume otherwise that has a rational solution . Then is rational if and only if the discriminant is a perfect square (using the quadratic formula, coupled with the fact that a,b,c are integers).

a,b,c are odd. We will show that cannot possibly be a perfect square by looking at it modulo 8. All of the odd perfect squares (1^2, 3^2, 5^2, 7^2) leave a remainder of 1 when divided by 8. However and , so , i.e. b^2 - 4ac always leaves a remainder of 5 when divided by 8. Therefore it cannot be a perfect square, and any real solution x cannot be rational.


Answer by ikleyn(52835)   (Show Source): You can put this solution on YOUR website!
.
Given odd integers a, b, c, prove that the equation cannot have a solution x which is a rational number.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Assume the equation  =  with odd integer coefficients a, b an c has the solution, 
which is a rational fraction  with integer p and q. 

We can assume that all the common divisors of p and q are just canceled in the fraction , 
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  into the equation.

You will get  = .

Multiply both sides by  to rid off the denominators. You will get

 = .   (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.


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