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They can't be odd!

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Prove that if a, b, and c are odd integers, then a x^2 + b x + c = 0 has no rational roots.

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For the roots to be irrational, the discriminant must not be a square number.

 

Using a proof by contradiction, assume that the roots can be rational. i.e. assume that the discriminant is a square number, k2

 

b2 - 4ac = k2.

b2 = 4ac + k2

 

Per the Pythagoren Theorem, 4ac must also be a square, which only occurs when a = c.

 

b2 - 4ac = k2

b2 - 4a2 = k2

 

Note that because b is odd, b2 is odd. And because 4a2 is even, k2, and therefore k, must also be odd. Let b = 2m + 1 and k = 2n+1

 

b2 - 4a2 = k2

(2m+1)2 - 4a2 = (2n+1)2

4m2 + 4m + 1 - 4a2 = 4n2 + 4n + 1

4m2 + 4m - 4a2 = 4n2 + 4n

m(m+1) - a2 = n(n+1)

 

Note that m(m+1) is even, a2 is odd, and n(n+1) is even. An odd number subtracted from an even number is always odd. We have reached a contradiction. Therefore, the discriminant can never be a squared number, and the roots can never be rational.

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Joyandwarmfuzzies wrote:

"Per the Pythagoren Theorem, 4ac must also be a square, which only occurs when a = c."

4ac may be a square other than when a = c. It is a square when all the factors of 4, a and c can be paired up (4 is is paired up with factors 2*2). As each "a" and "c" are given as odd, neither will have the prime factor 2 as a factor. Yet, "a" may have, for example, the factors 3*3*3 and "c" may simply have the factored 3. This would allow 4ac in this given example to be the square of 18 (=2*3*3).

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Turns out my note about a = c doesn't matter.

 

 

For the roots to be irrational, the discriminant must not be a square number.

 

Using a proof by contradiction, assume that the roots can be rational. i.e. assume that the discriminant is a square number, k2

 

b2 - 4ac = k2.

 

Per the Pythagorean Theorem, 4ac must also be a square. Note that because b is odd, b2 is odd. 4ac must be even. Therefore k2, and therefore k, must also be odd. Let b = 2m + 1 and k = 2n+1

 

b2 - 4ac = k2

(2m+1)2 - 4ac = (2n+1)2

4m2 + 4m + 1 - 4ac = 4n2 + 4n + 1

4m2 + 4m - 4ac = 4n2 + 4n

m(m+1) - ac = n(n+1)

 

Note that m(m+1) is even, ac is odd, and n(n+1) is even. An odd number subtracted from an even number is always odd. We have reached a contradiction. Therefore, the discriminant can never be a squared number, and the roots can never be rational.

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Turns out my note about a = c doesn't matter.

 

 

For the roots to be irrational, the discriminant must not be a square number.

 

Using a proof by contradiction, assume that the roots can be rational. i.e. assume that the discriminant is a square number, k2

 

b2 - 4ac = k2.

 

Per the Pythagorean Theorem, 4ac must also be a square. Note that because b is odd, b2 is odd. 4ac must be even. Therefore k2, and therefore k, must also be odd. Let b = 2m + 1 and k = 2n+1

 

b2 - 4ac = k2

(2m+1)2 - 4ac = (2n+1)2

4m2 + 4m + 1 - 4ac = 4n2 + 4n + 1

4m2 + 4m - 4ac = 4n2 + 4n

m(m+1) - ac = n(n+1)

 

Note that m(m+1) is even, ac is odd, and n(n+1) is even. An odd number subtracted from an even number is always odd. We have reached a contradiction. Therefore, the discriminant can never be a squared number, and the roots can never be rational.

 

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Oops... the reference to the Pythagorean Theorem is now irrelevant after making the change that DejMar noted, since 4ac being a square doesn't actually matter. I should have removed that, my mistake.

 

For the roots to be irrational, the discriminant must not be a square number.

 

Using a proof by contradiction, assume that the roots can be rational. i.e. assume that the discriminant is a square number, k2

 

b2 - 4ac = k2.

 

Note that because b is odd, b2 is odd. 4ac must be even. Therefore k2, and therefore k, must also be odd. Let b = 2m + 1 and k = 2n+1

 

b2 - 4ac = k2

(2m+1)2 - 4ac = (2n+1)2

4m2 + 4m + 1 - 4ac = 4n2 + 4n + 1

4m2 + 4m - 4ac = 4n2 + 4n

m(m+1) - ac = n(n+1)

 

 

Note that m(m+1) is even, ac is odd, and n(n+1) is even. An odd number subtracted from an even number is always odd. We have reached a contradiction. Therefore, the discriminant can never be a squared number, and the roots can never be rational.

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Proof by contradiction

Given ax^2 + bx +c = 0 and a,b,c are odd

Assume that x has a rational solution

Then there exist a p and q such that x = p/q where p/q is a reduced fraction and p,q are integers.

Then a(p/q)^2 + b(p/q) + c =0

Times everything by q^2

Ap^2 +bq +cq^2 = 0

Now we have three cases, both p and q are odd or one of them is odd.

1. Both are odd...

If both p and q are odd then ap^2, bq, and cq^2 are odd, meaning their sum is odd not zero.

2. p is odd ...

Then ap^2 is odd, while bq and cq^2 are even. And odd summed with two evens is odd not zero.

3. Q is odd...

The same reasoning follows if Q is odd, except you end up with an even plus and even plus and odd, which results in an odd not zero.

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