Suppose we have $ p, q$ where $ p$ and $q$ are twin primes between 6,10000. Now the following sequence $pq + 2$ generates 22 primes . These primes are all in the congruence class $ [1]_{100}$ I've been asked to prove why this is the case. I've spent a good time thinking about it and not got any ideas how to even start. Wondering if someone could give me just a direction at least to follow. Was considering the notion of parity as pq is always going to be odd.
Best regards
On
Rather than considering parity, you might want to consider the situation modulo $6$ to start with.
If our three primes are $p, p+2, p^2+2p+2$ we have $p\equiv -1 \bmod 6$ automatically and as observed in a comment on the question it is easy to check that we require $p\equiv -1 \bmod 5$ to avoid divisibility by $5$
We therefore have $p\equiv -1 \bmod 30$ and $p(p+2)+2\equiv 1 \bmod 900$
On
Hint: Consider in which cases the product of two twin primes (above $5$) doesn't end in $3$ (which would make their product plus $2$ end in $5$).
On
The comment by reuns holds the key to the answer. He points out that $p,p+2,p(p+2)+2$ being prime implies that $p\equiv -1 \mod 5$. But consider further that if $p\equiv -1 \mod 5$, then $p+2\equiv 1 \mod 5$, which means that the last digit of $p+2$ must be either $1,6$. But if the digit is $6$, then $p+2$ isn't prime. Therefore, taking odd/even into account, $p+2\equiv 1 \mod 10\Rightarrow p=10m-1$ and $p+2=10m+1$. Therefore $p(p+2)+2\equiv 1 \mod 100$ as you observe.
Look at the possible remainders $\bmod 10$. The only case where $p, p+2, p(p+2)+2$ may be all primes is $p\equiv 9\pmod {10} $ (otherwise there are factors $2$ or $5$). So we have $p=10q+9$ and then
$$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$