Let, $ N= (19 \cdot 29 \cdot 59...\cdot q)^{2} - 5.,$ where the primes numbers of the form $10n+9$ has been considered in the product. Then $N$ is of the form $20n-4$ and has odd prime factors in the form $10t \pm1$. Is it possible for $N$ to have only odd prime factors of the form $10k+1$?
I am having difficulty with this . It seems to be true while testing with some numbers by direct calculation. But I can't show it in general.
I would be highly grateful if someone tell me whether it is true or false. Any help would be appreciated. Thanks in advance.
let $q_1=19,$ $ q_2 = 29, $ and so on. Let $Q_r = q_1 q_2 \cdots q_r.$ As each $q_j \equiv -1 \pmod 5,$ $Q_r \equiv \pm1 \pmod 5,$ and $Q_r^2 \equiv 1 \pmod 5.$ For that matter, an odd square is 1 mod 8, so $Q_r^2 \equiv 1 \pmod {40}.$
For the first few $r$ we find $$ \frac{Q_r^2 - 5}{4} \equiv -1 \pmod 5. $$
Then $$ Q_{r+1}^2 - Q_r^2 \equiv 0 \pmod {40} $$
$$ \frac{Q_{r+1}^2 - Q_r^2}{4} \equiv 0 \pmod {10} \equiv 0 \pmod 5 $$
$$ \frac{Q_{r+1}^2 - 5}{4} = \frac{Q_{r+1}^2 - Q_r^2}{4} + \frac{Q_r^2 - 5}{4} $$
$$ \frac{Q_{r+1}^2 - 5}{4} = \frac{Q_r^2 - 5}{4} \equiv -1 \pmod 5 $$
by induction, this always holds, and $Q_w^2-5$ must always have at least one prime factor congruent to $-1 \pmod 5$