Question 23, Find all positive integers $a$, such that for any $n \in\{0,1,2, \ldots, a-1\}$, $4 n^{2}+a$ is a prime number.
Question 23:
Solution: When $n=0$, $4 n^{2}+a=a$ is a prime number.
Upon inspection, $\mathrm{a}=2$ does not satisfy the condition; $\mathrm{a}=3$ satisfies the condition; $\mathrm{a}=5$ does not satisfy the condition; $\mathrm{a}=7$ satisfies the condition. Now assume $\mathrm{a} \geq 11$, that is, $\mathrm{a}$ is a prime number no less than 11.
First, prove two lemmas.
Lemma 1: $a+1$ is a power of 2.
Proof of Lemma 1: If $\mathrm{a}+1$ has an odd prime factor $\mathrm{p}$, then $\mathrm{p} \leq \frac{\mathrm{a}+1}{2}$, so $\frac{\mathrm{p}+1}{2} \in\{0,1, \ldots, \mathrm{a}-1\}$, hence $4\left(\frac{p+1}{2}\right)^{2}+a$ is a prime number. But $4\left(\frac{p+1}{2}\right)^{2}+a=(p+1)^{2}+a \equiv 1+a \equiv 0(\bmod p)$, and $4\left(\frac{p+1}{2}\right)^{2}+a=(p+1)^{2}+a>p$, so $4\left(\frac{p+1}{2}\right)^{2}+a$ is a composite number, which leads to a contradiction. Therefore, $a+1$ is a power of 2.
Lemma 2: $a+9$ is a power of 2.
Proof of Lemma 2: If $\mathrm{a}+9$ has an odd prime factor $\mathrm{p}$, then $\mathrm{p} \leq \frac{\mathrm{a}+9}{2}$, so $\frac{\mathrm{p}+3}{2} \in\{0,1, \ldots, \mathrm{a}-1\}$, hence $4\left(\frac{p+3}{2}\right)^{2}+a$ is a prime number. But $4\left(\frac{p+3}{2}\right)^{2}+a=(p+3)^{2}+a \equiv 9+a \equiv 0(\bmod p)$, and $4\left(\frac{p+3}{2}\right)^{2}+a=(p+3)^{2}+a>p$, so $4\left(\frac{p+3}{2}\right)^{2}+a$ is a composite number, which leads to a contradiction. Therefore, $a+9$ is a power of 2.
Now, return to the original problem.
According to the above analysis, $\mathrm{a}$ is a prime number no less than 11, $a+1$ is a power of 2, and $a+9$ is a power of 2. Let $\left\{\begin{array}{l}a+1=2^{\alpha} \\ a+9=2^{\beta}\end{array}\right.$, then $2^{\alpha}+2^{3}=2^{\beta}$. Therefore, it can only be $\left\{\begin{array}{l}\alpha=3 \\ \beta=4\end{array}\right.$, thus $\mathrm{a}=2^{3}-1$. This contradicts $\mathrm{a} \geq 11$.
In conclusion, the values of $\mathrm{a}$ that satisfy the condition are 3 or 7.
Note: In fact, for any odd prime $\mathrm{p}<\mathrm{a},\left\{4 \mathrm{n}^{2} \mid \mathrm{n} \in\{0,1,2, \ldots, \mathrm{a}-1\}\right\}$ can cover all quadratic residues modulo $\mathrm{p}$, so for any square number $b^{2}, a+b^{2}$ cannot have an odd prime factor less than $a$ (otherwise, if $a+b^{2}$ has an odd prime factor $p$ less than $a$, then there exists an integer $n \in\{0,1,2, \ldots, a-1\}$, such that $p \mid\left(4 n^{2}+a\right)$, thus $4 n^{2}+a$ cannot be a prime number). Therefore, when $b^{2}<a$, and $b^{2}+a$ is composite, $b^{2}+a$ can only be a power of 2. This is the lemma used in the above proof.