I'm working on a homework problem showing that some elements of $\mathbb{Z}[\sqrt{10}]$ are irreducible but not prime. I'm using the norm function that maps $a+b\sqrt{10}$ to $a^2-10b^2$. After proving several other results, I conjecture that $a^2-10b^2\neq\pm 2, \pm 3$ for any integers $a$ and $b$. If my conjecture holds, all of my other work falls into place. I've been trying to prove this for a few hours and I'm stumped at this point. I'd prefer not to be told the answer, but any helpful hints (or a counterexample to my conjecture so that I can try something else) would be greatly appreciated. I don't have a ton of number theory to work with, so I need to use a pretty elementary direction for proving it.
Show that $a^2-10b^2\neq \pm 2, \pm 3$
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abstract-algebra
number-theory
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1$a^2$ cannot be $\pm 2$ or $\pm 3 \mod 10$. – 2017-01-18
1 Answers
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After the hint given in the comments, it is clear that $a^2-10b^2=\pm 2,\pm 3 \Rightarrow a^2\equiv \pm 2, \pm 3 \pmod{10}$.
One can then observe that any integer $a$ is of the form $10x+y$ with $x$ and $y$ integers and $0 \leq y \leq 9$.
Some minor work shows $a^2 \equiv y^2 \pmod{10}$ so it suffices to show that $y^2$ is not congruent to $\pm 2, \pm 3$ modulo 10, which can be done case by case.