$a$ and $b$ are two integers such that: $$(1+\sqrt{2})^{2020} = a + b\sqrt{2}$$
What is the value of: $a^2 - 2b^2$? And I know that $a^2-2b^2=(a+b\sqrt 2)(a-b\sqrt 2)$ but the problem is Idk the value of $a-b\sqrt 2$ Note that newton's binomials or anything like that is not allowed as I didn't study it,yet it's still dedicated to my level. Thx in advance for everyone who's trying to help
Let $$ a_n+b_n\sqrt2=(1+\sqrt{2})^{n} $$ and hence $$ a_{n+1}+b_{n+1}\sqrt2=(1+\sqrt{2})^{n+1}=(a_n+b_n\sqrt2)(1+\sqrt2). $$ So $$ a_{n+1}=a_n+2b_n,b_{n+1}=a_n+b_n,a_1=b_1=1. $$ Clearly $a_2=3,b_2=2,a_3=7,b_3=5$. Observe $$ a_1^2-2b_1^2=-1, a_2^2-2b_2^2=1, a_3^2-2b_3^2=-1 $$ and hence you can conjecture $$ a_n^2-2b_n^2=(-1)^n. $$ Now you can use Mathematical Induction to prove it. So $$ a^2-2b^2=1. $$
So here's an approach lets consider the map $t(x): \mathbb{Z}[\sqrt{2}] \rightarrow \mathbb{Z}[\sqrt{2}]$ given by $t(x+y\sqrt{2}) = x-y\sqrt{2}$. Translated into casual speak this is a function which accepts something of the form $x+y\sqrt{2}$ and outputs $x-y\sqrt{2}$. So for example $t(1+\sqrt{2}) = 1-\sqrt{2}$. $t(5-2\sqrt{2}) = 5+2\sqrt{2}$ etc...
Now unless you have experience with these things, you might be quite surprised to know that $t(a+b) = t(a) + t(b)$ and $t(ab) = t(a)t(b)$ (you can check this manually, and i'll add it to the answer in a bit). A corollary of the second identity is that $t(a^2) = t(a)*t(a) = t(a)^2$ and more generally $t(a^n) = t(a)^n$
It then follows that if
$$ (1+\sqrt{2})^{2020} = a + b\sqrt{2}$$
Then:
$$ t((1+\sqrt{2})^{2020}) = t(a + b\sqrt{2})$$
So we then have on the left hand side:
$t((1+\sqrt{2})^{2020}) = t(1+\sqrt{2})^{2020} = (1-\sqrt{2})^{2020}$
and on the right hand side:
$t(a+b\sqrt{2}) = a-b\sqrt{2}$
So putting it all together, if
$$ t((1+\sqrt{2})^{2020}) = t(a + b\sqrt{2})$$
Then:
$$ (1-\sqrt{2})^{2020} = a-b\sqrt{2} $$
Now recall the observation that
$$a-2b^2 = (a-b\sqrt{2})(a+b\sqrt{2})$$
So then it must be that
$$ a- 2b^2 = (1-\sqrt{2})^{2020} (1+\sqrt{2})^{2020} = (1 - 2)^{2020} = (-1)^{2020} = 1$$
For the advanced reader: we say that $t(x)$ is a Ring Automorphism of $\mathbb{Z}[\sqrt{2}]$ and more specifically a Ring Automorphism that fixes $\mathbb{Z}$. In simpler terms "it's the same kind of thing as complex conjugation".
Consider the binomial expansion of $\left(1+\sqrt2\right)^{2020}$. $$\left(1+\sqrt2\right)^{2020}=\sum_{r=0}^{2020}{2020\choose r}\left(\sqrt2\right)^r=\sum_{k=0}^{1010}{2020\choose2k}2^k+\sqrt2\sum_{m=0}^{1009}{2020\choose2m+1}2^m$$ Therefore, $a=\sum_{k=0}^{1010}{2020\choose2k}2^k$ and $b=\sum_{m=0}^{1009}{2020\choose2m+1}2^m$.
As @JakeMirra suggested, also consider the expansion of $\left(1-\sqrt2\right)^{2020}$. $$\left(1-\sqrt2\right)^{2020}=\sum_{r=0}^{2020}{2020\choose r}\left(-\sqrt2\right)^r=\sum_{k=0}^{1010}{2020\choose2k}2^k-\sqrt2\sum_{m=0}^{1009}{2020\choose2m+1}2^m$$ so $\left(1-\sqrt2\right)^{2020}=a-b\sqrt2$.
As mentioned in the original post, $$a^2-2b^2=\left(a+b\sqrt2\right)\left(a-b\sqrt2\right)=\left(1+\sqrt2\right)^{2020}\left(1-\sqrt2\right)^{2020}=(-1)^{2020}=\boxed{1}$$
If direct methods like binomial theorem are not allowed, we can try to do it by induction. The logic is simple, we want to prove the following statement is true: If $(1+\sqrt{2})^{n} = a+b\sqrt{2}$, then $(1-\sqrt{2})^{n} = a-b\sqrt{2}$.
First we check when $n=1$, the result is trivial.
Now assume the statement is true for $n=k$, we have for $n=k+1$, $$(1+\sqrt{2})^{k+1} = (1+\sqrt{2})^{k}(1+\sqrt{2}) = (a+b\sqrt{2})(1+\sqrt{2}) \\= (a+2b)+(a+b)\sqrt{2} = a'+b'\sqrt{2},$$ where $ a' =(a+2b), b' = (a+b)$. Now let's look at the other part of the statement, $$(1-\sqrt{2})^{k+1} = (1-\sqrt{2})^{k}(1-\sqrt{2}) = (a-b\sqrt{2})(1-\sqrt{2}) \\= (a+2b)-(a+b)\sqrt{2} = a'-b'\sqrt{2},$$ the result follows.
Thus for $n = 2020$, we have $(1+\sqrt{2})^{2020} = a+b\sqrt{2}$ and $(1-\sqrt{2})^{2020} = a-b\sqrt{2}$. So $$a^2-2b^2 = (1+\sqrt{2})^{2020}(1-\sqrt{2})^{2020} = (1^2-(\sqrt{2})^2)^{2020} = 1$$
There are a lot of Correct & good Answers already.
One thing missing is the requirement given by OP :
"... newton's binomials or anything like that is not allowed ... I didn't study it ... dedicated to my level"
I assume OP can not use Induction / Maps / rings / automorphisms / matrix operations ...
Here is one Simple way :
$$X=(1+\sqrt{2})^{2020} = a + b\sqrt{2}$$
We can calculate $1/X$ easily. $$\frac{1}{X}=\frac{1}{(1+\sqrt{2})^{2020}} = \frac{1}{[a + b\sqrt{2}]}$$
We want to rationalize the Denominator ...
$$\frac{1}{X}=\frac{(1-\sqrt{2})^{2020}}{(1+\sqrt{2})^{2020}(1-\sqrt{2})^{2020}} = \frac{[a - b\sqrt{2}]}{[a + b\sqrt{2}][a - b\sqrt{2}]}$$
$$\frac{1}{X}=\frac{(1-\sqrt{2})^{2020}}{(1^2-\sqrt{2}^2)^{2020}} = \frac{[a - b\sqrt{2}]}{[a^2 - b^2\sqrt{2}^2]}$$
$$\frac{1}{X}=\frac{(1-\sqrt{2})^{2020}}{(1-2)^{2020}} = \frac{[a - b\sqrt{2}]}{[a^2 - 2b^2]}$$
When $a$ & $b$ have no common factors , we will have only Integers in the middle term Expansion ( observation only ... not necessary to use binomial theorem ) , hence the last term Expansion must also have only Integers , hence $[a^2 - 2b^2]=\pm{1}$
We can then see that we have Positive $X$ , hence Positive $[a^2 - 2b^2]$
Thus $[a^2 - 2b^2]=1$
I am using the idea of Jake Mirra in the comments above, so I thank him first. If you know about basic Galois theory, then (using the fact $\mathbb{Q}[\sqrt{2}]$ is a extension of $\mathbb{Q}$, and the map $\phi: \mathbb{Q}[\sqrt{2}] \rightarrow \mathbb{Q}[\sqrt{2}]$ is automorphism), note if $a+b\sqrt{2}=(1+\sqrt{2})^{2020}$ then (applying $\phi$) we note $a-b\sqrt{2}=(1-\sqrt{2})^{2020}$. Now, as already mentioned $a+b\sqrt{2}=(1+\sqrt{2})^{2020}$ and $a-b\sqrt{2}=(1-\sqrt{2})^{2020}$ should give the value of $a^2-2b^2$.
This article should explain Galois theory, if you need:
