p and q are odd numbers, prove that the quadratic equation
$x^2+px+q=0$ has
1) no integer solution
2) no rational solution

Question

p and q are odd numbers, prove that the quadratic equation
$x^2+px+q=0$ has
1) no integer solution
2) no rational solution

amistre64 · Answer

use the quadratic formula ....

amistre64 · Answer

$$x=\frac{-(2k+1)\pm\sqrt{(2k+1)^2-4(2k+1)(2n+1)}}{2(2n+1)}$$

amistre64 · Answer

i misplaced my p and q ...

amistre64 · Answer

$$x=\frac{-(2k+1)\pm\sqrt{(2k+1)^2-4(1)(2n+1)}}{2}$$

ujjwal · Answer

i guess, basically we need to prove that the discriminant is never a positive perfect square number.
Don't know if that's a correct approach..

amistre64 · Answer

yes, and recall that k,n in Z

amistre64 · Answer

if:$$4k^2+4k-8n-8+1\ge0$$
then there is a chance for it, so as long as we can show the other stuff

notice that is an odd number to start with ..... not that there arent odd perfect squares, but it might reduce the work load

ujjwal · Answer

yep.
and it's $4k^2+4k-8n-4+1\geq 0$
and 9 is an odd number and a perfect square.
are there no other odd perfect squares?

amistre64 · Answer

15, 49, 81,  odd^2 = odd

4(2n-1) >= 4k^2 + 4k + 1

2n-1 >= k^2 + k + 1/4

ujjwal · Answer

squares of all odd numbers are odd

amistre64 · Answer

yep

amistre64 · Answer

2n+1 >= k^2 + k + 1/4

2n + 1 >= (k +1/2)^2

n >= [(k +1/2)^2 - 1] /2

ujjwal · Answer

if that's the case, the discriminant is positive.
but we don't even know if that's the case or not or any such thing..

ujjwal · Answer

I don't get it much..

amistre64 · Answer

do we have to try a direct approach?  or can we use contradiction or contrapositive?

ujjwal · Answer

I don't have much idea..
Any which way, this has to be proved..

anonymous · Answer

try to show that the discriminant of quadratic can not be a perfect square

ujjwal · Answer

exactly.. but i don't know how?

amistre64 · Answer

if sqrt(p^2-4q) is odd, then we have an integer

if sqrt(p^2-4q) is a perfect square, then we have a rational

ganeshie8 · Answer

see if it works :-

say,

$\large x^2+px+q = (x-\alpha)(x-\beta)$

comparing coeffecients both sides, 

$\large \alpha + \beta = -p$ ----(1)

$\large \alpha  \beta = q$ ----- (2)

substitute $\beta = -p- \alpha$ from (1) in (2)


$\large -\alpha(p + \alpha)   = q$ 

right hand side is a odd, but left hand side is even. contradiction - so integer roots are not possible

amistre64 · Answer

a+b is odd if a and b are odd

odd(odd+odd) = odd ?

odd(even) = odd ?

2n(2k+1) = 2(2nk+n) is even

ujjwal · Answer

i get it @ganeshie8 . Thanks!
But then what about the irrational thing?

amistre64 · Answer

3+1 = 4

3 + 3 = 6

odd + odd = even

ujjwal · Answer

i mean the second question which is asking to prove that the solution is not rational.

amistre64 · Answer

im outta time for today ... good luck

ganeshie8 · Answer

oh, second q looks tricky... @mukushla

ujjwal · Answer

$\alpha +\beta =p$ which is odd,
so, among $\alpha$ and $\beta$ one is odd while the other is even.
but then,
$\alpha\beta=q$
left side is even, right side is odd
contradictory

ganeshie8 · Answer

u made it look much simpler :)

ganeshie8 · Answer

### amistre lines above :

if sqrt(p^2-4q) is odd, then we have an integer

if sqrt(p^2-4q) is a perfect square, then we have a rational

should work ?

anonymous · Answer

oh guys, i was out...let$$\Delta=p^2-4q=n^2$$$n$ is odd, let $p=2k+1 , \ n=2m+1$$$(p-n)(p+n)=4q$$$$(k-m)(k+m+1)=q$$left hand side is even and right hand side is odd
hence both parts of question proved

ujjwal · Answer

Thanks a lot!
You made it a lot simpler!

anonymous · Answer

very welcome bro :)