P1 and P3 are number theory?

Mathological03 wrote: P1 and P3 are number theory? 5 problems in total for 5 hours

Let $a_i=x+i-1$, so $2(a_1+\cdots+a_k)+a_1-a_1^2=(2x+k-1)k+x-x^2=-x^2+x(2k+1)+k(k-1)$. The discriminant of $(-x^2+x(2k+1)+k(k-1))-2$ is $(2k+1)^2+4(k^2-k-2)=8k^2-7$ which is a square infinitely often. Therefore, for infinitely many $k$, there exists an $x$ such that $2(a_1+\cdots+a_k)+a_1-a_1^2=2$ which divides $2(a_1\cdots a_{k-2}-1)^2$.

If we take $k = p \geq 2024$ and $a_1 = p+1$; with $p$ being prime, then we get the set $A=\{p+1, p+2, ..., 2p\}$. In the numerator by wilson's throrem, we get: $$\prod_{i=1}^{k-2}a_i \equiv \prod_{i=1}^{p-2} i = (p-2)! \equiv 1 \pmod{p}$$$$\iff \text{numerator} \equiv 0 \pmod{2p^2}$$Now for the denominator: $$2 \times (a_1 + a_2 + \dots + a_k) - a_1^2 + a_1 = 2 k \times a_1 + 2 \times \frac{k(k-1)}{2} - a_1^2 + a_1$$$$\iff \text{denominator} = 2p(p+1) + p(p-1) - (p+1)^2 + (p+1) = 2p^2$$This suffices as there are infinitely many primes greater than 2024 and for each of them, the quantity is an integer. Remark: the denominator is just twice the quantity, $$\sum_{i=n+1}^{2n} i - \sum_{i=1}^{n} i= n ^2$$Which is always a perfect square, We just have n=p.