# Prove that there are no integer solutions to $3m^2-1=n^2$.

Prove that it is impossible to find any integer $n$ such that $n^2 \equiv 2 \pmod 4$ or $n^2 \equiv 3 \pmod 4$. Hence or otherwise, prove that there do not exist integers $m$ and $n$ such that $3m^2 - 1 = n^2$.

I'm still stuck :(

Regarding to "what is n^2 congruent to each of the 0,1,2,3 cases, is it just 0,1,4,9? What do I do next?

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Is this homework? What have you tried? – Bruno Joyal May 10 '12 at 18:38

It is also impossible to find any integer $n$ such that $$n^2 \equiv 2 \pmod 3.$$

EDITTTTT: Note that if you have a positive prime $p \equiv 1 \pmod 4,$ then there is always an integer solution to $$x^2 - p y^2 = -1.$$ Proof in Mordell's book. For prime coefficients there is thus no ambiguity, as for positive prime $q \equiv 3 \pmod 4,$ there is never an integer solution to $$x^2 - q y^2 = -1.$$

It starts to get tricky with composite coefficients. There is no integer solution to $x^2 - 34 y^2 = -1.$ There is no integer solution to $x^2 - 205 y^2 = -1.$ Here 205 is the smallest odd number that gives a surprise. And, when I say surprise, note that there is never a solution to $x^2 - k^2 y^2 = -1$ when $k >1.$

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HINTS:

1. An integer $n$ must be congruent to one of $0,1,2$, and $3$ mod $4$. To what is $n^2$ congruent in each of these cases?

2. After you've proved the first statement, you know that $n^2$ must be congruent to $0$ or $1$ mod $4$, so $n^2+1$ must be congruent to $1$ or $2$ mod $4$, and of course $3\equiv -1\pmod 4$. What can you deduce from this about $m^2$ mod $4$?

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Hint $\rm\: mod\ 4\!:\ even^2\equiv 0,\ odd^2\equiv 1\:\Rightarrow\ n^2\! + m^2\not\equiv 3,\$ i.e. $\rm\ n^2\! - 3\:\!m^2\not\equiv -1\$

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