Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As it says in the headline, how does one prove that there doesn't exist $a, b$ $\in \Bbb Z$ with $|a^2-5b^2|=2$? Any hints?

share|cite|improve this question

put on hold as off-topic by user26857, USER91500, Shailesh, Watson, Claude Leibovici 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, USER91500, Shailesh, Watson, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

up vote 4 down vote accepted

$5b^2\pm2\equiv \pm2\pmod 5$

$a$ can be of the form $5k,5k\pm1,5k\pm2$ where $k$ is any integer.

But $(5k)^2\equiv 0\pmod 5,(5k\pm1)^2\equiv 1\pmod 5, (5k\pm2)^2\equiv 4\equiv -1\pmod 5,$

So, $a^2\equiv 0,\pm1\pmod 5\not\equiv\pm2$

Observation: $2\equiv-3, -2\equiv 3$ are quadratic non-residues of $5$, unlike $1\equiv -4,-1\equiv 4$.

share|cite|improve this answer
    
I was just about to write this :) +1 – Belgi Nov 19 '12 at 5:45

HINT: Suppose that $|a^2-5b^2|=2$; then either $a$ and $b$ are both odd, or they’re both even. (Why?) If they’re both even, $a^2-5b^2$ is a multiple of $4$. (Why?) If $a$ is odd, then $a^2$ is either a multiple of $5$ or one more than a multiple of $5$. (Why?)

share|cite|improve this answer

For this to be true, you would need either:

$a^2 - 5b^2 = 2$ or $a^2 - 5b^2 = -2$.

Reducing modulo $5$, the first equation becomes $a^2 = 2$ mod $5$.

Similarly, the second equation becomes $a^2 = 3$ mod $5$.

But any square reduced modulo $5$ must be $0, 1,$ or $4$.

So neither equation can hold for integers $a$ and $b$.

share|cite|improve this answer

Hint $\rm\ \ (a\!-\!b)(a\!+\!b) = 4b^2 \pm 2\:\Rightarrow\:2\mid a\pm b\:\Rightarrow\:4\mid 2\:\Rightarrow\Leftarrow$

share|cite|improve this answer

Hint: For any $x$, we have $x^2\equiv 0\pmod{5}$ or $x^2\equiv 1\pmod{5}$.

share|cite|improve this answer
1  
Per chance you mean : for any odd $x$? because 2² satisfies not your conclusion... I mean : take $x=2$. – awllower Nov 19 '12 at 5:47

$n^2 \in \{0, 1\}$ ($\bmod$ $4$) for all integral $n$ and $x^2 - 5y^2 \equiv x^2 - y^2$ ($\bmod$ $4$). The difference of two squares can only be $-1 (\equiv 3), 0$ or $1$ ($\bmod$ $4$).

share|cite|improve this answer

First show that $a$ and $b$ must be relatively prime to $2$, then show that $2$ and $-2$ are not quadratic residues modulo $5$. So one concludes that $\mid a^2-5b^2\mid=2$ has no solution in integers.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.