SOLUTION: If a, b, c are odd integers, then the equation {{{ ax^2+bx+c=0 }}} has no FRACTION solution. I know how to prove, if it said integer solution, but to prove that theres no fract

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Question 1017072: If a, b, c are odd integers, then the equation has no FRACTION solution.
I know how to prove, if it said integer solution, but to prove that theres no fraction solution, it seems a little hard for me.
Im really sorry if its the wrong section
Thanks~!
Found 2 solutions by ikleyn, richard1234:
Answer by ikleyn(52800)   (Show Source): You can put this solution on YOUR website!
.
If a, b, c are odd integers, then the equation has no FRACTION solution.
I know how to prove, if it said integer solution, but to prove that theres no fraction solution, it seems a little hard for me.
Im really sorry if its the wrong section
Thanks~!
-------------------------------------- 

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.


Answer by richard1234(7193)   (Show Source): You can put this solution on YOUR website!
You probably mean *rational* solution (since "fraction" is somewhat ambiguous).

The roots of the equation are . It suffices to prove that cannot be a perfect square, otherwise the solutions would be rational.

If is a perfect square, then where a, b, c are odd integers and n is an integer. Note that n is odd, since b^2 is odd and 4ac is even.

Here, we use a little modular arithmetic. The right hand side must leave a remainder of 4 when divided by 8, since a and c are odd. However, all of the odd squares leave a remainder of 1 when divided by 8, meaning that their difference is a multiple of 8. Since the remainders upon division by 8 are not equal, the expressions and cannot possibly be equal, and there is no solution in odd integers a,b,c. Therefore cannot be a perfect square, no rational solution.

In modular arithmetic terms, we say that and .
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