How many integers in the range satisfy a given congruence?

Question

Specifically how many integers $x$ in the range $1 ≤ x ≤ 60$ satisfy $21x ≡ 24\ (mod\ 60)$? And how many integers $a$ are there in the range $1 ≤ a ≤ 74$ for which the equation $x^2 ≡ a\ (mod\ 37)$ is soluble?

I have the solutions to be $3$ and $38$ respectively but not too sure how they came up with those answers.

For the first one, I believe it's because $21x ≡ 24\ (mod\ 60) \implies 3\cdot7x \equiv 3\cdot8\ (mod\ 60) \implies 7\cdot8x\ (mod\ 20) \implies x \equiv 4\ (mod\ 20)$ Then looking in the range from $1\ to\ 60$, $60/20 = 3$ to give us 3 integers.

Not sure about the second..

Answer 1

For the second one, it's basically asking how many quadratic residues are in $Z_{37}$. Since 37 is prime, there must be exactly $\frac{p-1}{2}$ residues and non-residues in the range from $1-36$, now, 37 is a trivial residue, so you would get $18+1=19$ residues in the range $1\leq a\leq37$ As we are in modulo 37, the range $38\leq a \leq74$ behaves similarly as $1\leq a\leq37$. So you would get $19*2=38$ residues in total.

Answer 2

\begin{align} 21x &\equiv 24 \pmod{60} \\ 7x &\equiv 8 \pmod{20} \\ x &\equiv 4 \pmod{20} \\ x &\in \{4, 24, 44 \} \end{align}

TOTAL = 3

---

\begin{array}{|c|c|} \hline x & x^2 \pmod{37}\\ \hline 0,37 & 0 \\ 1,36 & 1 \\ 2,35 & 4 \\ 3,34 & 9 \\ 4,33 & 16 \\ 5,32 & 25 \\ 6,31 & 36 \\ 7,30 & 12 \\ 8,29 & 27 \\ 9,28 & 7 \\ 10,27 & 26 \\ 11,26 & 10 \\ 12,25 & 33 \\ 13,24 & 21 \\ 14,23 & 11 \\ 15,22 & 3 \\ 16,21 & 34 \\ 17,20 & 30 \\ 18,19 & 28 \\ \hline \end{array}

TOTAL = $2 \times 19 = 38$