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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?

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    Please tell us what kind of proof you are talking of. – 2017-02-20
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    I'm not sure what you mean? We were just given this proof as a homework and we can use any method. So far we've learned direct, contradiction, and both induction methods. – 2017-02-20
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    Any method? Is mine OK? – 2017-02-20
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    Yes, yours makes perfect sense! Thank you for your help! – 2017-02-20
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    @MPW thank you for the edit, I'm fairly new to LaTeX and was unsure how to to do subscripts, so thank you for the additional knowledge! – 2017-02-20
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    @Deez1133 : No problem. Note that you can right-click on math expressions and select "Show Math As TeX Commands" to see what generated them. You can browse around and look under the hood that way. – 2017-02-20

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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 $3$.

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    _Right, that's about it..._ – 2017-02-20
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    @ParclyTaxel What exactly do you mean by this? – 2017-02-20
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    It means: this is the simplest possible proof, and I was about to type it in, but then you came. – 2017-02-20
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    Okay! This makes sense! Thank you so much! – 2017-02-20
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    The simplest one. +1 – 2017-02-20
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    @Xam Thanks for your upvote. – 2017-02-20
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    @S.C.B I fully plan on it, but it said I had to wait 4 minutes to do so. So once that time is up it will be accepted. – 2017-02-20
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    @Deez1133 As a general rule for the future, and no offence meant to SCB (+1), I advise against very rapid acceptance of answers. It reduces the number of people looking at your question (and indeed looking at the answers) so you might miss out on another good insight. – 2017-02-20
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    @Joffan, understood. That makes sense and you are correct. Thank you! – 2017-02-20
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    @Joffan I didn't know that either. Thanks. For some reason I wasn't notified by your comment. Wierd... – 2017-02-20
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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 $a^2-2b^2=-1$.

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@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 $2t+1$.

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    Ah, but does $3b^2-1$ have $2t+1$ factors? $74=3 \times 5^2-1$ has only two factors. – 2017-02-20
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    @S.C.B. I answered the question the OP had (I think). I don't know what proof he was referring to, but it was "similar" -- not the same. I don't think that argument could apply here, so I would tell the OP to use your proof instead. – 2017-02-20
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    @scott you answered my question for sure, I understand how the conclusion was made now, thank you for that! – 2017-02-20