# Modular Arithmetic. Integer Solutions

How can you show that $$|a^2 -10b^2|=2$$ has no integer solutions for a and b using modular arithmetic? Thank you.

Modulo $$5$$, we get $$a^2\equiv 2\mod 5$$ or $$a^2\equiv 3\mod 5$$ which both is not solvable over the integers.

• Thank you. Do you get these two options by considering the two possible (positive and negative options) from the absolute value sign? Also, why are both not solvable over the integers. Is it because a^2 = 4 mod 5 or a^2 = 0 mod 5? Thank you. – Maths May 28 at 17:23
• $(1)$ yes, the absolute value gives the possibilities $a^2=2$ $(mod\ 5)$ or $a^2=-2=3$ $(mod\ 5)$ $(2)$ , we have $a^2=0,1$ or $4$ $mod\ 5$, so $2$ and $3$ are impossible. – Peter May 28 at 17:32
• Thanks you. Please could you mention how you we have 𝑎^2=0,1 or 4 𝑚𝑜𝑑 5? Is there a general fact for x^2 and modulo arithmetic. Thank you. – Maths May 28 at 17:46
• @Maths The easiest way is to simply go through the possible values for a mod 5, which are 0,1,2,3,4. A more difficult approach is the theory of the quadratic residues (quadratic non-residues) modulo a number (in particular modulo a prime). – Peter May 28 at 17:49
• Thanks. Is it true that x^2 =0(mod 4) or x^2=1mod(4) for every integer x. Therefore, does it follow that x^2 =4(mod 5) or x^2 =0(mod 5) – Maths May 28 at 17:53

This is a case where the principal form also represents $$-1,$$ as $$3^2 - 10 = -1.$$ So, if $$a^2 - 10 b^2 = n,$$ we find $$(3a+10b)^2 - 10 (a+3b)^2 = -n$$

Meanwhile, $$x^2 - 10 y^2$$ represents primes $$p \equiv 1, 9, 31, 39 \pmod {40}$$

The other class of this discriminant $$2x^2 - 5 y^2,$$ represents $$2$$ and $$5,$$ then primes $$p \equiv 3, 13, 27, 37 \pmod {40}$$

If $$2a^2 - 5 b^2 = k,$$ we find $$2(3a+5b)^2 - 5 (2a+3b)^2 = -k$$

Best to work modulo $$10$$. You know, from elementary school, that the only squares modulo $$10$$ are $$0,1,4,5,6$$, and $$9$$. The impossibility of $$\pm2=m^2-10n^2$$ follows.