Prove any odd square cannot be of the form $4n+3$

Question

This is both a problem, and an opportunity for me to learn proof building, so sorry for being too detailed.

An 'odd square number' means : a square number which is odd also, e.g. $9, 25, 49, 121, 169$.
Square of an odd number (which can be of the form : $2n+1$) will be of the form $4n^2 + 4n+1= 4n(n+1)+1=8n+1$.

The original statement ($P\implies Q$) is:
An odd square number cannot be of the form $4n+3$ for some integer $n$:
$P$ : An odd square number;
$Q$: $\exists n \in \mathbb{Z}$, s.t. form of the number is $4n+3$,
with the problem being represented as: $P \implies \neg Q$.

But, unable to directly prove the title.

Only an indirect proof can be attempted by me.
If a given statement is: $P\implies Q\equiv \neg P \vee Q$, then using indirect approach, $4$ different hypothetical approaches are:
(i) Proving by converse : if $Q$, then $P$ => $\neg Q\vee P$. But, there is no relation between truth value of converse and the original, as also given here.
(ii) Proving by inverse : if not $P$, then not $Q$ => $\neg P\implies \neg Q$, or $P \vee \neg Q$. But there is no relation between truth value of inverse and the original, as given in link at (i).
(iii) Proving by contrapositive : if not $Q$, then not $P$ => $\neg Q\implies \neg P $, or $Q \vee\neg P \equiv$ the original statement.
(iv) Proving by contradiction (reductio ad absurdum): My approach is based on article here, that differentiates clearly between negation & contradiction.
Prove falsity of: $\neg (P\implies Q)$ => $\neg (\neg P \vee Q)$ => $P \wedge \neg Q$

Let me attempt each of the two approaches, i.e. (iii), (iv).

(i) contrapositive based approach: Need prove the falsity of: $\neg(\neg Q) \implies \neg P\equiv Q \implies \neg P$ :
$ \exists n \in \mathbb{Z}$ such that the form of number is $4n+3 \implies $ An even square number.
Unable to prove above, but it is easy to prove falsity of the converse of the above, i.e. : an even square number is of the form $4n+3$.

(ii) contradiction based approach: Here, am unable to even formulate the problem correctly. Or if it is correct, then am unable to pursue.
The problem here is:
Prove that $\neg (P\implies \neg Q)\equiv \neg(\neg P \vee \neg Q)\equiv P\wedge Q$ is false, i.e. to prove that the below combination is false:
(An odd square number)$\wedge (\exists n \in \mathbb{Z}$ such that the form of number is $4n+3)$

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Edit Have two observations:
1. It is given by answers that the contrapositive approach needs the help of 'informal' contradiction, i.e. showing that contrapositive is false.
The selected approach shows in a more rigorous (by which, I mean (i) a more formal way, (ii) a more clearer way, that relies less on words and more on showing the truth/falsity, to explain to any one else by me) manner.
2. Also, need show that the contradiction approach is false in a similar manner.

Can there be a better 'word' for 'informal' usage of the contradiction, so as to make it more easy to understand or frame proofs? I wish that word should not be negation, as that also has a formal meaning, as shown by a link above, as a technique distinct from contradiction.
Why not the word 'falsity/false/falsehood' is universally used, to show this 'informal' negation/contradiction?

Answer 1

Your owring is very confusing. For instance

Proving by converse : if Q, then P => ¬Q⟹P, or Q∨P, or P∨Q. But, there is no relation between truth value of converse and the original, as also given here.

You have "if .. then", => and ⟹. Are you using them to mean different things? Does "if Q, then P => ¬Q⟹P" mean "( if Q, then P) => (¬Q⟹P)"? or "if Q, then (P => ¬Q⟹P)" or something else?

As for the actual proof, we have either k = 2n or k=2n+1. If k =2n, then $k^2=4n^2$, which is even, so this is not an odd square. So we have to have k = 2n+1. This part can be viewed as proof by contradiction/contrapositive; if we square an even number, we don't get an odd number, so if we have an odd number, we couldn't have gotten it by squaring an even number.

Once we have that k = 2n+1, we have that $k^2 = 4n^2+4n+1=4(n^2+n)+1$, so $k^2$ is equal to 1 modulus 4. As for now showing it's not equal to 3 modulus 4, this follows directly from the properties of modulus; modulus is a unique number, so if its modulus is 1, then its modulus is not 3. But if you really want a rigorous proof, you could do a contrapositive, $k^2 = 4n'+3$. Since $k^2 = 4(n^2+n)+1$, we have $4(n^2+n)+1=4n'+3$ and $4(n^2+n-n')=3-1=2$, therefore $n^2+n-n'=\frac 2 4 = \frac 1 2$. But n^2+n-n' must be an integer, which is a contradiction.

Answer 2

Let observe that

$$(2n+1)^2\equiv 1 \pmod 4$$

but

$$4n+3\equiv 3 \pmod 4$$

Answer 3

It might be useful to prove: if $x^2$ is odd then $x$ is odd.

Then if $x^2$ is odd, the $x$ is odd, then $x$ is of the form $2m + 1$ and $x^2 = 4m^2 + 4m + 1 = 4(m^2 + m) + 1$ and is of the form $4n + 1$, and therefore is not of the form $4n+3$.

This assumes:

1) All odd numbers are of the form $2m + 1$.

2) All numbers are either of the form $4n; 4n + 1; 4n+2; 4n+3$ and the those are exhaustive and mutually exclusive.

3) And it assumes if $x^2$ is odd then $x$ is odd.

Answer 4

Suppose on the contrary that $n$ is an odd square and $n \equiv 3 \pmod{4}$.

That is suppose $\exists m \in \mathbb{Z}$, $n=m^2$ and $n \equiv 1 \pmod{2}$ and $n \equiv 3 \pmod{4}$.

We can consider two cases:

Case $1$: $m$ is an even number, that is $\exists k \in \mathbb{Z}, m=2k$, then $n=(2k)^2=4k^2 \equiv 0 \pmod{4}$ which contradicts that $n \equiv 3\pmod{4}$. Hence we rule out this case.

Case $2$: $m$ is an odd number, that is $\exists k \in \mathbb{Z}, m=2k+1$, then $n=(2k+1)^2=4(k^2+k)+1 \equiv 1 \pmod{4}$ which contradicts that $n \equiv 3 \pmod{4}$.

Hence the assumption that $n$ is an odd square and $n \equiv 3 \pmod{4}$ must be wrong.