(a)Show that a product of numbers of the form $4k+1$ also have this form.
Easy peasy: Using modular arithmetic: $a_1 a_2 \cdots a_n = 1\cdot 1\cdots1 = 1\ \pmod 4$.
(b)Deduce that if $n=-1 \pmod 4$ and $n>0$ then $n$ must have a prime factor $q=-1 \pmod 4$.
By the fundamental theorem of arithmetic $n$ can be written as a product of primes. I said that every prime is of the form $4k-1$ or $4k+1$ since $4k+2$ and $4k$ cannot possibly be primes (they are divisible by $2$) and $4k+3 = 4(s+1)-1$. But from above a product of number of the form $4k+1$ have the form $4k+1$. So n must have one prime factor of the form $4k-1$.
(c) Show that for $m\geq 4$, $n=m!-1$ has a prime factor $q=-1\pmod 4$ and $q>m$. Deduce that the primes of the form $4k-1$ are infinite.
$n=m!-1$ means $n=-1\pmod 4$ since $4$ is a factor of $m!$ for $m\geq4$. The prime factor part follows from part (b). I had some trouble proving $q>m$. I tried a contradiction argument: Suppose $q\leq m$. Then got $m≥1$ and $m≥4$. Contradiction? Or $m=2$ gives no prime factors. I'm guessing the infinite bit follows from $q>m$?
(d) What happens if we replace 4 by 6 or by 8 in parts a,b,c. I said the same results apply but I am not sure.