Answer :
Step-by-step explanation:
To verify if the congruence $x^2 \equiv 10 \pmod{89}$ is solvable, we can use the Legendre symbol. The Legendre symbol $(\frac{a}{p})$ for an odd prime $p$ and integer $a$ is defined as:
- $(\frac{a}{p}) = 1$ if $a$ is a quadratic residue modulo $p$ (i.e., the congruence $x^2 \equiv a \pmod{p}$ has a solution).
- $(\frac{a}{p}) = -1$ if $a$ is a quadratic non-residue modulo $p$.
- $(\frac{a}{p}) = 0$ if $a \equiv 0 \pmod{p}$.
We need to calculate $(\frac{10}{89})$. According to the law of quadratic reciprocity, if $p$ and $q$ are odd primes, then:
$(\frac{p}{q}) \cdot (\frac{q}{p}) = (-1)^{\frac{(p-1)(q-1)}{4}}$
We can use this to find $(\frac{10}{89})$ by breaking it down as $(\frac{2}{89}) \cdot (\frac{5}{89})$.
First, let's find $(\frac{2}{89})$. For $p$ an odd prime, $(\frac{2}{p}) = (-1)^{\frac{p^2-1}{8}}$. Since $89 \equiv 1 \pmod{8}$, we have:
$(\frac{2}{89}) = (-1)^{\frac{89^2-1}{8}} = (-1)^{\frac{7920}{8}} = (-1)^{990} = 1$
Now, let's find $(\frac{5}{89})$. We use the law of quadratic reciprocity:
$(\frac{5}{89}) \cdot (\frac{89}{5}) = (-1)^{\frac{(5-1)(89-1)}{4}} = (-1)^{\frac{4 \cdot 88}{4}} = (-1)^{88} = 1$
Since $89 \equiv 4 \pmod{5}$, we have:
$(\frac{89}{5}) = (\frac{4}{5}) = (\frac{2^2}{5}) = 1$
So, $(\frac{5}{89}) = 1$.
Combining both, we get:
$(\frac{10}{89}) = (\frac{2}{89}) \cdot (\frac{5}{89}) = 1 \cdot 1 = 1$
Since $(\frac{10}{89}) = 1$, the congruence $x^2 \equiv 10 \pmod{89}$ is solvable. There exists an integer $x$ such that $x^2$ is congruent to $10$ modulo $89$.