First, we wish to show that
A
∪
(
B
∩
C
)
⊂
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle A\cup (B\cap C)\subset (A\cup B)\cap (A\cup C)}
. Let
x
∈
A
∪
(
B
∩
C
)
{\displaystyle x\in A\cup (B\cap C)}
. Then
x
∈
A
{\displaystyle x\in A}
or
x
∈
B
∩
C
{\displaystyle x\in B\cap C}
.
Case 1:
x
∈
A
{\displaystyle x\in A}
x
∈
A
⊂
A
∪
B
{\displaystyle x\in A\subset A\cup B}
so that
x
∈
A
∪
B
{\displaystyle x\in A\cup B}
x
∈
A
⊂
A
∪
C
{\displaystyle x\in A\subset A\cup C}
so that
x
∈
A
∪
C
{\displaystyle x\in A\cup C}
x
∈
A
∪
B
{\displaystyle x\in A\cup B}
and
x
∈
A
∪
C
{\displaystyle x\in A\cup C}
so that
x
∈
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle x\in (A\cup B)\cap (A\cup C)}
Case 2:
x
∈
B
∩
C
{\displaystyle x\in B\cap C}
x
∈
B
{\displaystyle x\in B}
and
x
∈
C
{\displaystyle x\in C}
x
∈
B
⊂
A
∪
B
{\displaystyle x\in B\subset A\cup B}
so that
x
∈
A
∪
B
{\displaystyle x\in A\cup B}
x
∈
C
⊂
A
∪
C
{\displaystyle x\in C\subset A\cup C}
so that
x
∈
A
∪
C
{\displaystyle x\in A\cup C}
x
∈
A
∪
B
{\displaystyle x\in A\cup B}
and
x
∈
A
∪
C
{\displaystyle x\in A\cup C}
so that
x
∈
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle x\in (A\cup B)\cap (A\cup C)}
Since in both cases,
x
∈
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle x\in (A\cup B)\cap (A\cup C)}
, we know that
A
∪
(
B
∩
C
)
⊂
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle A\cup (B\cap C)\subset (A\cup B)\cap (A\cup C)}
Now we wish to show that
(
A
∪
B
)
∩
(
A
∪
C
)
⊂
A
∪
(
B
∩
C
)
{\displaystyle (A\cup B)\cap (A\cup C)\subset A\cup (B\cap C)}
. Let
x
∈
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle x\in (A\cup B)\cap (A\cup C)}
. Then
x
∈
A
∪
B
{\displaystyle x\in A\cup B}
and
x
∈
A
∪
C
{\displaystyle x\in A\cup C}
.
Case 1a:
x
∈
A
{\displaystyle x\in A}
x
∈
A
⊂
A
∪
(
B
∩
C
)
{\displaystyle x\in A\subset A\cup (B\cap C)}
, so
x
∈
A
∪
(
B
∩
C
)
{\displaystyle x\in A\cup (B\cap C)}
Case 1b:
x
∈
B
{\displaystyle x\in B}
We can't actually conclude anything we want with just this, so we have to also to consider the case
x
∈
A
∪
C
{\displaystyle x\in A\cup C}
.
Case 2a:
x
∈
A
{\displaystyle x\in A}
: [see Case 1a]
Case 2b:
x
∈
C
{\displaystyle x\in C}
We now have
x
∈
B
{\displaystyle x\in B}
and
x
∈
C
{\displaystyle x\in C}
so that
x
∈
B
∩
C
{\displaystyle x\in B\cap C}
Of course, since
x
∈
B
∩
C
⊂
A
∪
(
B
∩
C
)
{\displaystyle x\in B\cap C\subset A\cup (B\cap C)}
, it follows that
x
∈
A
∪
(
B
∩
C
)
{\displaystyle x\in A\cup (B\cap C)}
. Since both cases 2a and 2b yield
x
∈
A
∪
(
B
∩
C
)
{\displaystyle x\in A\cup (B\cap C)}
, we know that it follows from 1b.
Since in both cases 1a and 1b,
x
∈
A
∪
(
B
∩
C
)
{\displaystyle x\in A\cup (B\cap C)}
, we know that
(
A
∪
B
)
∩
(
A
∪
C
)
⊂
A
∪
(
B
∩
C
)
{\displaystyle (A\cup B)\cap (A\cup C)\subset A\cup (B\cap C)}
.
Since both
A
∪
(
B
∩
C
)
⊂
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle A\cup (B\cap C)\subset (A\cup B)\cap (A\cup C)}
and
(
A
∪
B
)
∩
(
A
∪
C
)
⊂
A
∪
(
B
∩
C
)
{\displaystyle (A\cup B)\cap (A\cup C)\subset A\cup (B\cap C)}
, it follows (finally) that
A
∪
(
B
∩
C
)
=
(
A
∪
B
)
∩
(
A
∪
C
)
{\displaystyle A\cup (B\cap C)=(A\cup B)\cap (A\cup C)}
.
--will continue later, feel free to refine it if you feel it can be--
Because the question asks about the square of a number, you can substitute the definition of an odd number 2n + 1 into the number to be squared. So, say x is that number, then
x
2
=
(
2
n
+
1
)
2
=
(
2
n
+
1
)
(
2
n
+
1
)
{\displaystyle x^{2}=(2n+1)^{2}=(2n+1)(2n+1)}
Multiply both factors together
(
2
n
+
1
)
(
2
n
+
1
)
=
4
n
2
+
4
n
+
1
{\displaystyle (2n+1)(2n+1)=4n^{2}+4n+1}
Factor out a two for the first two terms
4
n
2
+
4
n
+
1
=
2
(
2
n
2
+
2
n
)
+
1
{\displaystyle 4n^{2}+4n+1=2(2n^{2}+2n)+1}
The factor
2
n
2
+
2
n
{\displaystyle 2n^{2}+2n}
will always be a natural number. As such, it fits the definition of an odd number, 2n + 1
Problem solved!