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I am going through the Book of Proof by Richard Hammack, and in the Chapter on Proof by contradiction, I came across the following exercise:

Prove that if $a,b\in\mathbb{Z}$, then $a^{2}-4b-3\neq 0$ by contradiction.

My solution went on a different aproach as the solution provided in the end of the book, so I would like some help to evaluate if it is correct, and tips for improvement.

My proof was as following:

For the sake of contradiction, suppose $a,b\in\mathbb{Z}$ and $a^{2}-4b-3= 0$. Hence, $a^{2}-3=4b$, which means $a^{2}\equiv 3(\textrm{mod }4)$.

Note that $a$ is either even or odd.

Suppose $a$ is even. Thus, there is an integer $k_1$ such that $a=2k_1$. So, $a^{2}=4k_1^{2}$, which means $a^{2}\equiv 0(\textrm{mod }4)$, a contradiction. Hence, $a$ is not even. Therefore, $a$ is odd.

Since $a$ is odd, there is an integer $k_2$ such that $a=2k_2+1$. Hence, $a^{2}=(2k_2+1)^{2}=4k_2^2+4k_2+1=4(k-2^{2}+k_2)+1$. Thus $a^{2}\equiv 1(\textrm{mod }4)$, a contradiction. Thus, $a^{2}-4b-3\neq 0$. $\blacksquare$

Is this correct? Any tips on improving?

Igor Mello
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    This is correct. By curiosity, what is the approach in the book ? Yours seems the most natural to me. – TheSilverDoe Jun 27 '22 at 17:29
  • Begin with showing that the choices for a square are $x^2 \equiv 0,1 \pmod 4$ Call it a lemma if you like – Will Jagy Jun 27 '22 at 17:36
  • A "fancier" way of saying the intermediate facts you noted is "the quadratic residues modulo $4$ are $0$ and $1$". – aschepler Jun 27 '22 at 17:37
  • This is correct. We can shave just a few words off by noting for that as $a^2\equiv 3 \pmod 4$ that means $a^2$ is odd so $a$ is odd and we don't need to consider $a$ even. But your contradiction is just as good (actually better as my way requires showing that any $3+4b$ is odd; and showing that $a$ even $\implies a^2$ is even. Both are frivilous and can be "done in ones head" and I would grant they go without saying but others might not.) – fleablood Jun 27 '22 at 18:07
  • I'd also like to see how your book did it. I imagine although it may on the surface look different is probably in essence and spirit the same as yours. – fleablood Jun 27 '22 at 18:10

2 Answers2

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Alternative way:

Let $a^2-4b-3=0$, then

$a^2-1=4b+2$

$(a-1)(a+1)=2(2b+1)$

Both $a-1$ and $a+1$ have the same parity, both are odd or even.

Case 1: Both are odd, but RHS is even

Case 2: Both are even, but RHS is not a multiple of $4$.

Since in both cases, there is a contradiction, $a^2-4b-3\neq0$, $a,b \in Z$

Lion Heart
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Let's assume a² - 4b - 3 = 0 when a and b are integers.
a² - 3 = 4b
if a is even left side of this equation is odd but right side is even .
if a is odd let a = 2n +1 where n is an integer.
(2n+1)² - 3 = 4b ⇒ n(n+1) - 1/2 = b , this result is not valid because left side is not an integer , therefore by the method of contradiction a² - 4b - 3 ≠ 0 when a , b are integers.

Janaka Rodrigo
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