# Math Help - Elementary Number Theory

1. ## Elementary Number Theory

Prove the equation $x^2 = 4y + 3$ has no integer solutions.

So, I was thinking, prove by contradiction that statement is not true. So:
Assume $x^2 = 4y+ 3$ has integer solutions.

This is where I get stuck. Do you let x and y be some integers k and j? How do you prove (or rather, disprove) integer solutions??

2. If a number x is even, then it is of the form 2n and so its square is 4n^2 and so can not be in the form 4y + 3.

If a number x is odd, then it is of the form 2n+1 and so its square is ...?

Can that form of x be of the form 4y + 3?

3. OK - you lost me with the not written in 4y+3 form.
Why would $4k^2$ have to look like 4y+3 : y and k are not the same variables ... ?

4. y is an integer. k^2 is an integer. 4k^2 is not in the same form as 4y+3 because it's 4y + 3 is 3 more than a multiple of 4. 4k^2 is zero more than a multiple of 4.

Have you covered modulo arithmetic?

5. Hello, cassiopeia1289!

Prove the equation $x^2 \:= \:4y + 3$ has no integer solutions.
Assume there exist integers $x$ and $y$ that satisfy the equation.

We have: . $x^2 \:=\:4y+3 \quad\Rightarrow\quad y \:=\:\frac{x^2-3}{4}$

$x$ must either even or odd

If $x$ is even, then $x \:= \:2k$ for some integer $k.$

. . Then: . $y \;=\;\frac{(2k)^2-3}{4} \;=\;\frac{4k^2-3}{4}\;=\;k^2 -\frac{3}{4}$

Hence, $y$ cannot be an integer.

If $x$ is odd, then $x \:= \:2k+1$, for some integer $k.$

. . Then: . $y \;=\;\frac{(2k+1)^2 - 3}{4} \;=\;\frac{4k^2 + 4k - 2}{4} \;=\;k^2+k - \frac{1}{2}$

Hence, y cannot be an integer.

Therefore, no integer solution exists.

6. Originally Posted by Soroban
Hello, cassiopeia1289!

Assume there exist integers $x$ and $y$ that satisfy the equation.

We have: . $x^2 \:=\:4y+3 \quad\Rightarrow\quad y \:=\:\frac{x^2-3}{4}$

$x$ must either even or odd

If $x$ is even, then $x \:= \:2k$ for some integer $k.$

. . Then: . $y \;=\;\frac{(2k)^2-3}{4} \;=\;\frac{4k^2-3}{4}\;=\;k^2 -\frac{3}{4}$

Hence, $y$ cannot be an integer.

If $x$ is odd, then $x \:= \:2k+1$, for some integer $k.$

. . Then: . $y \;=\;\frac{(2k+1)^2 - 3}{4} \;=\;\frac{4k^2 + 4k - 2}{4} \;=\;k^2+k - \frac{1}{2}$

Hence, y cannot be an integer.

Therefore, no integer solution exists.

Thank you - thank you - thank you!
I get that.

7. ok sorry - same part of homework: new question(s)

"If you get at least a C on the final exam, then you will pass the course." So if you got a D on the final - then what can you conclude?

He didn't pass OR there's still hope

OK: and second one, similar in nature:

"If you get at least a B on the final, then you will pass the course" You pass the course. What can you conclude?

He got at least a B on the final OR you can't conclude anything
(On this one, I'm inclined to think you can't conclude anything, because P does not cause Q, so he could still pass some other way.)