Thank you for sharing the image. I see several mathematical problems in this document. I'll analyze and solve them one by one. If you have a specific question you'd like to start with, please let me know. Otherwise, I’ll begin with Q2 (B, part c). Let me know your preference!

If you're ready, here are the first few problems in detail:

Q2 (B, c): Find the least positive integer $x$ such that $13 \mid (x^2 + 1)$

This means $x^2 + 1$ must be divisible by $13$. Formally, we solve: $x^2 + 1 \equiv 0 \pmod{13}$ $x^2 \equiv -1 \pmod{13}$ x^2 \equiv 12 \pmod{13} \quad \text{(since -1 \equiv 12 \mod{13})}.

Now, we need to find the values of $x$ for which $x^2 \equiv 12 \pmod{13}$. Let’s test the integers from $x = 0$ to $x = 12$ (the modular residue classes):

• $x = 0: 0^2 = 0 \not\equiv 12 \pmod{13}$,
• $x = 1: 1^2 = 1 \not\equiv 12 \pmod{13}$,
• $x = 2: 2^2 = 4 \not\equiv 12 \pmod{13}$,
• $x = 3: 3^2 = 9 \not\equiv 12 \pmod{13}$,
• $x = 4: 4^2 = 16 \equiv 3 \pmod{13} \not\equiv 12$,
• $x = 5: 5^2 = 25 \equiv 12 \pmod{13} \quad \text{(solution found)}$.

Thus, the least positive integer $x$ is: $\boxed{5}.$

Would you like me to proceed with the next sub-question or expand on this? Let me know!

Related Questions:

1. What is the general solution for quadratic congruences like this one?
2. How can we check for modular inverses efficiently?
3. What is the modular arithmetic property of $(-1)^2$?
4. How to calculate quadratic residues for arbitrary primes?
5. What if we change $13$ to another prime?

Tip: Always check modular congruences systematically using quadratic residue theory for efficiency.