# Proofs (1 Viewer)

#### Gtsh

##### Member
How do I do Q10? Thanks

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#### cossine

##### Active Member
How do I do Q10? Thanks
Are you having trouble with 10 i) or 10 ii)

Please show working out.

#### Gtsh

##### Member
I'm stuck for part ii). For part i) I put "If the equation a^2x+bx+cx has at least one integer solution, then either a,b or c is even"
I'm not too sure if it's right or not.

#### Masaken

##### Active Member
I'm stuck for part ii). For part i) I put "If the equation a^2x+bx+cx has at least one integer solution, then either a,b or c is even"
I'm not too sure if it's right or not.
I *think* you would start by basically putting in the quadratic formula:

then since x is divided by 2a, and x is an integer, the numerator must be even and = 2xa (which proves that at least b, a or c is even). i'm not too sure though

#### cossine

##### Active Member
I'm stuck for part ii). For part i) I put "If the equation a^2x+bx+cx has at least one integer solution, then either a,b or c is even"
I'm not too sure if it's right or not.
Part i) is correct however I would avoid the use of either or. This because wither or can be taken mean exclusive or.

#### hogzillaAnson

##### New Member
The contrapositive is given by
$\bg_white \text{If}\ ax^2+bx+c=0\ \text{has an integer solution in}\ x,\ \text{then at least one of}\ a,b,c\ \text{is even.}$

Let's prove the contrapositive. We know from the quadratic formula that

\bg_white \begin{align*} x &= \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ 2ax &= -b\pm \sqrt{b^2-4ac} \\ (\text{even}) &= -b\pm \sqrt{b^2-4ac} \end{align*}
$\bg_white \therefore b\ \text{and}\ \sqrt{b^2-4ac}\ \text{both odd or both even}$
$\bg_white \text{However, if}\ b\ \text{is odd, then}\ b^2\ \text{is odd, and}\ b^2-4ac=(\text{odd})-(\text{even})=\text{odd},\ \text{a contradiction}$
Thus it must be the case that b is even. That concludes our proof