1. The problem statement, all variables and given/known data Show that the number of primes of the form 4n-1 and 4n +1 are infinite 2. Relevant equations 3. The attempt at a solution I am able to show this for 4n -1 but I am having trouble doing it for primes of the other form. ( I am hoping to do it without using modular arithmetic) For primes of the form 4n-1 I used the same argument that was used in Euclid proof by showing that I can always generate a new prime of that form : Let 4a-1, 4b-1, 4c-1,...,4k-1 be all the primes of the above form let x = 4[(4a-1)(4b-1)(4c-1)...(4k-1)] -1 1. x has a prime divisor that is not part of the above list because if it was it would have to divide 1 which would make it not prime 2. the prime divisor is odd so it is of the form 4y+1 or 4y-1 3. because (4y+1)(4z+1) = 16yz + 4z + 4y + 1 = 4(4yz+y+z) +1 , not all the prime divisors can be of the form 4y+1, one must be of the form 4y-1 so there is a prime divisor that can always be generated in this manner therefore there are an infinite number of primes of the form 4n-1. When i try to prove there are an infinite number of primes of the form 4n+1 and I get to the part ( 3. from previous proof) I get (4y-1)(4z-1)= 16yz -4y -4z +1 =4(yz -y -z) +1 which doesn't tell me that all the factors are not of the form 4y-1 so i can't make the conclusion that one must be of the form 4y +1. What other way should I approach this.