First, we wish to show that
. Let
. Then
or
.
Case 1:
so that ![{\displaystyle x\in A\cup B}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43b649b2f9459be716c3d7d15cfbac8f27ce3cd4)
so that ![{\displaystyle x\in A\cup C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0386ca018f888670e5601dfab8f9b733cedd1ad)
and
so that
Case 2:
and ![{\displaystyle x\in C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7fc788379bff7289bdf694ffd68ee690e999eb7)
so that ![{\displaystyle x\in A\cup B}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43b649b2f9459be716c3d7d15cfbac8f27ce3cd4)
so that ![{\displaystyle x\in A\cup C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0386ca018f888670e5601dfab8f9b733cedd1ad)
and
so that
Since in both cases,
, we know that ![{\displaystyle A\cup (B\cap C)\subset (A\cup B)\cap (A\cup C)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d47e4ed9dad0a7f0f7f897e51482bffd98d298c6)
Now we wish to show that
. Let
. Then
and
.
Case 1a:
, so
Case 1b:
We can't actually conclude anything we want with just this, so we have to also to consider the case
.
Case 2a:
: [see Case 1a]
Case 2b:
We now have
and
so that ![{\displaystyle x\in B\cap C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2589e1f7f0aae397183a469eeeaf5ee833a86dbb)
Of course, since
, it follows that
.
Since both cases 2a and 2b yield
, we know that it follows from 1b.
Since in both cases 1a and 1b,
, we know that
.
Since both
and
, it follows (finally) that
.
--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
![{\displaystyle x^{2}=(2n+1)^{2}=(2n+1)(2n+1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cb21e5c9d20ca2018769f9bb23e7636cc8f19df)
- Multiply both factors together
![{\displaystyle (2n+1)(2n+1)=4n^{2}+4n+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/968702547fcb630949cb9faf9d0f4e8256fc69ef)
- Factor out a two for the first two terms
![{\displaystyle 4n^{2}+4n+1=2(2n^{2}+2n)+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59fc21c1f24c57d2f25281f3c49d4490418ecea8)
- The factor
will always be a natural number. As such, it fits the definition of an odd number, 2n + 1
- Problem solved!