Mathematics Number Theory Proof: Prove that there are no integers $a$ and $b$ such that: $a^2-3b^2 = -1$

Question

Prove that there are no integers $a$ and $b$ such that $a^2-3b^2 = -1$.

I got a hint to use prime factorization, so I rearrange the equation to be $a^2 = 3b^2-1$ and set $a$ to be $a= p_1p_2\cdots p_n$ and $b$ to be $b=t_1 t_2\cdots t_m$.

I saw a similar question online where they concluded that there are $2n$ primes on the left side of the equation and $2t+1$ primes on the right, and since euclid's fundamental theorem states there is only one factorization, then we cannot have even=odd, hence the proof is concluded.

But I do not understand how it was concluded that there are $2n$ and $2t+1$ primes on either side of the equation?

Answer 1

I think this is what we can do $$n^2 \equiv 0,1 \pmod 3$$For all $n$. But $$a^2=3b^2-1 \equiv -1 \pmod {3}$$ Contradiction, as a square number can only be $0$ or $1$ modulo $3$.

Answer 2

For every $n\in\mathbb{N}$, we have $n^2\equiv0\pmod{3}$ or $n^2\equiv1\pmod{3}$.

If $a,b$ are integers such that $a^2-3b^2=-1$ then $a^2\equiv2\pmod{3}$ : a contradiction.

Note that there exist (infinitely many) integers $a,b$ such that $a^2-3b^2=1$.

This question is bound to the determination of invertible elements in the ring $\mathbb{Z}[\sqrt 3]$.

By contrast, it's interesting to notice that there exist (infinitely many) integers solutions to both diophantine equations $a^2-2b^2=1$ and $a^2-2b^2=-1$.

Answer 3

@S.C.B. has an elegant proof, but to answer the question about the proof you found online:

If $a$ has $n$ prime factors, then $a^2$ has $2n$ factors. The same is true for $b$: if $b$ has $t$ factors, then $b^2$ has $2t$ factors. But the equality is $3b^2$, so the total number of factors on that side of the equation is $2t+1$.