1. ## Prime Numbers!

In this question arithmetic in a restricted subset of Z, similar to arithmetic with E-numbers,
is investigated. Let D = {4a + 1 | a ∈ Z} and call the elements of D the D-numbers. The
first few positive D-numbers are

1
, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57.

Put another way, the D-numbers are the integers which are congruent to 1 modulo 4. Use
of arithmetic modulo 4 makes answers to some of the the following questions very simple
(but is not mandatory).

(a) Show that if two D-numbers are multiplied together the result is a D-number. Give an
example to show that the same is not true when two D-numbers are added together.

(b) If a and b are D-numbers we say that a D-divides b if b = ac, where c is a D-number.
Show that 5 D-divides 25 and 45. Show that 1 D-divides every D-number.

(c) Now show that if a and b are D-numbers such that a|b (in the usual sense of Definition
1.5) then a D-divides b. Give an example to show that there are integers n and m,
which are not both D-numbers but are such that mn is a D-number.

(d) If a is a positive D-number greater than 1 and the only positive D-divisors of a are
1
and a , then we say that a is D-prime. List the first 10 D-primes and the first two
positive D-numbers (> 1) which are D-composite (i.e. not D-prime).

(e) All (ordinary) odd primes are either of the form 4m+1 or 4m+3: that is are congruent
to 1 or 3 modulo 4. Show that a D-number is D-prime if and only if it is prime (in Z)
or its prime factorisation is pq where p and q are congruent to 3 mod 4 (and may be
equal).

(f) Find a D-number which has two distinct D-prime factorisations.

2. Which question do you need help with?

3. Well for part (a), am i just to do (4a+1)*(4a+1) and then somehow show that this always = a multiple of (4a+1)?

4. Originally Posted by sirellwood
Well for part (a), am i just to do (4a+1)*(4a+1) and then somehow show that this always = a multiple of (4a+1)?
Not (4a+1)(4a+1) but two numbers of that form- the numbers multipying "4" may be different. $(4a+1)(4b+1)= 16ab+ 4(a+b)+ 1= 4( ? )+ 1$. That's pretty easy, isn't it?

What is (4a+ 1)+ (4b+1)? In particular, what is 5+ 9? Is it a "D number"?

5. Haha, yes it is very easy! yes i was able to do the 2nd part of part (a) :-)

For part (b) i have started off by saying that 5 = (4a+1) and 25 = (4b+1) and then hopefully going to say that (4b+1)/(4a+1) = a D-number?

Is this the right route?

6. I have also just tried this for (b) but not sure it is sufficient?

5=(4a+1)
25 = 5(4a+1)

$\frac{25}{5}$ = $\frac{5(4a+1)}{(4a+1)}$

= 5

= D-Number

Then a similar idea for 45