Prime sequence. 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
 A: 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 $5$).
A: 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.
A: 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 
$$p(p+2)+2 = (10q+9)(10q+11)+2 = 100q^2+200q+101$$

Therefore $p(p+2)+2 \equiv 1 \pmod {100}$

A: 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 $p(p+2)+2\equiv 1 \bmod 900$
