# 3

Let r=a+b\sqrt{2} and s=c+d\sqrt{2} \hspace{5pt} (a,b,c,d \in \mathbb{Z}).

\therefore rs = (ac+2bd) + (ad+bc)\sqrt{2}

\begin{align} \implies N(rs)&=(ac+2bd)^2-2(ad+bc)^2 \\ &=a^2c^2+4abcd+4b^2d^2-2a^2d^2-4abcd-2b^2c^2 \\ &=a^2(c^2-2d^2)-2b^2(c^2-2d^2) \\ &=(a^2-2b^2)(c^2-2d^2) \\ &=N(r)N(s) \end{align}

The equation a^2-2b^2=1 is effectively N(r)=1, where r=a+b\sqrt{2}\hspace{5pt} (a,b \in \mathbb{Z}). We can use the property that N(rs)=N(r)N(s) to prove that there are infinitely many integers satisying N(r)=1.

Notice that 1^n=1 for any n \in \mathbb{Z}. Therefore if we know one solution (a,b) for which N(r)=1 then N(r^2)=N(r)N(r)=1^2=1. Similarly, N(r^3)=1,N(r^4)=1 and so on.

It is not difficult to spot that (a,b)=(3,2) is a solution since 3^2-2(2)^2=1. Hence, it is also true that N\left((3+2\sqrt{2})^n \right)=1 for all n \in \mathbb{N}. Since the set \mathbb{N} is an infinite set, there exist infinitely many pairs of integers satisfying a^2-2b^2=1.

A similar thing can be done for the equation a^2-2b^2=-1 or N(r)=-1. We use the fact that (-1)^n=-1 for all odd numbers n. A simple solution is (a,b)=(1,1) since 1^2-2(1)^2=-1. So, it is also true that N\left((1+\sqrt2)^n \right)=1 for all odd numbers n. Since the set of odd numbers is an infinite set, there exist infinitely many pairs of integers satisfying a^2-2b^2=-1.
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Now we consider the equation a^2-3b^2=\pm1. Let’s redefine N(r)=a^2-3b^2 for all real numbers r of the form r=a+b\sqrt{3}\hspace{5pt} (a,b \in \mathbb{Z}). We also have N(rs)=N(r)N(s).

Let’s first take the case of a^2-3b^2=1. We can easily find a pair of integers (a,b) which is (2,1). Therefore, since we can find one pair, there must be infinitely many pairs, by the same logic with the previous case of a^2-2b^2=1.

Taking the second case of a^2-3b^2=-1, there isn’t an immediately obvious pair of integers that satisfy this. Let’s rearrange it to 3b^2=a^2+1, which implies that a^2+1 must be divisible by 3, if a,b \in \mathbb{Z}. However, a^2+1 can never be divisible by 3. Here’s a quick proof of this:

Write a=3q+r with r=0, r=1 or r=2 and (n,q \in \mathbb{Z}).
If r=0, a^2+1=3(3q^2)+1.
If r=1,a^2+1=3(3q^2+2q)+2.
If r=2,a^2+1=3(3q^2+4q+1)+2.
So, a^2+1 is never divisible by 3, and hence a^2-3b^2=-1 has no integer solutions.

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nicely done dude XD

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