# Prime pleasing an offered problem

wonderful one.
For $\frac{p-1}{4}$ to be weird, $p$ has to be of the kind $8k+5$, so $\frac{p-1}{4}$ is of the kind $2k+1$ and also $\frac{p+1}{2}$ of the kind $4k+3$.

If $k = 0,1,2 (\mod 3)$ after that the 3 numbers are specifically conforming to

$k=0(\mod 3): 2,1,0 (\mod 3)$

$k=1(\mod 3): 1,0,1 (\mod 3)$

$k=2(\mod 3): 0,2,2 (\mod 3)$ This suggests that the only means all 3 of them are prime numbers is that of them is $3$. For $k=0$ we have $5,1,3$ which is dismissed ; for $k=1$ we have $13,3,7$ which pleases theories ; for $k>1$ all numbers are $> 3$.
The only various other instance to be examined is $\frac{p-1}{4} = 2$ ; in this instance nonetheless $p = 9$, so this can not be a remedy.

The tops have item $(p^3 - p)/8$ which is divisible by $3$. So one prime = $3$. The remainder is unimportant.

The instances $p=2,3$ can be trivially examined. I'll think $p\ge 5$.

Keep in mind $\left( \displaystyle\frac{p-1}{4} \right) \left( \displaystyle\frac{p+1}{2}\right)=\displaystyle\frac{p^2-1}{8}=a$ (say). After that $a$ has just 2 prime divisors.

Currently it is popular that $24|p^2-1$ for $p>3$. Allow $p^2-1=24t$. After that $a=3t$.

Hence $3$ is a prime divisor of $a$, indicating $3$ amounts to among $\displaystyle\frac{p-1}{4},\frac{p+1}{2}$. Straight replacement reveals that $p=13$ is the only remedy.

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