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.
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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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