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Introduction

In this coursework I was asked to investigate the Phi Function (φ) of a number (n). The Phi Function of a number (n) is delineated as the number of positive integers less than n, which have no factor (other than 1) in common, i.e. co-prime with n.  Example: φ(16) = 8. The integers less than 16 that have no factors apart from 1 in common with 16 are 1, 3, 5, 7, 9, 11, 13, and 15. There are 8 altogether. To calculate φ(n) I will list out the numbers from 1 till n1. I will then cross out all the numbers that have a common factor with n. The remaining numbers will give me the φ of n.

In this part of the coursework will be investigating the phi function of:

1. φ(p)
2. φ(p)²

Part 1:

Find the value of:

1. φ(3):
1 2

3 = 1,2

The number 3 only has 2 positive co-prime integers they are the numbers 1 and 2.

(ii) φ(8):

1 2 3 4 5 6 7

8 = 1,3,5,7

There are 4 positive co-prime integers for the number 8

(iii) φ(11):

1 2 3 4 5 6 7 8 9 10 11

11 = 1,2,3,4,5,6,7,8,9,10

The number 11 has 10 positive co-prime integers, they are shown above.

(iv) φ(24):

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

24 = 1,5,7,11,13,17,19,23

The number 24 has 8 positive co-prime integers, they are shown above.

I will create a table for the Phi values of the numbers from 1 to 30. I have created this table because it would be easier to lookup the phi values of the numbers within this range instead of solving it every time I need to find a value quickly.

Middle

22

29

28

From the above tables, and the value tables, I have observed the following:

1. The phi values of the numbers as a whole do not need to increase as the number increases.
φ(8) = 4, φ(7) = 6
But for even numbers, the
φ values of an even number which is also a multiple of 3 or a multiple of 5 the value is less than that of the previous value, therefore we can say that:
φ(e, where e is an even integer and not a multiple of 5 or 3) < φ(the next e)
All the even numbers which are multiples of 3 or 5 break the increase of
φ values as the number increases.
φ(34) = 16, φ(36) = 12
Also for the odd numbers, the phi of a number o, where o is an odd number but not a prime number, increases as the value of n increases. We can say that:
φ(o, where n is odd but not prime) < φ(the next o).
The value of o increases as the odd number increases.
Similarly in prime numbers, the phi values of a number, p, were p is prime, the values of phi p increase as the number increases. We can say that:
φ(p) < φ(next p)
2. I have also found out the phi value of a number p where p is a prime number to be:
φ(p) = p – 1
This is logically true as all prime numbers have only 1 as a common factor, therefore all the numbers less that p will be co-prime to it. Check:
φ(3) = 2
φ(5) = 4
φ(29) = 28
Therefore we can say that the phi value of 199 =
φ(199) = 198
3. I have also found out that the phi value of a number o, where o is an odd number, = the phi values of the number double it. Therefore we can say that:
φ(o) = φ(2 x o)
Check:
φ(3) = φ(6)
φ(3) = 2, φ(6) = 2
φ(3) is equal to φ(6)
φ(5) = 4, φ(10) = 4
φ(7) = 6, φ(14) = 6
Therefore we can say that:
φ(27) = 18 = φ(54)
4.  I have also observed that to find the value of the phi of an even number, you could find that by checking the phi of its double, by twice of its answer. I.e. the phi of a number e, were e is an even number, is equal to the phi of twice that number divided by 2. We can see that:
2 x (
φ(e)) = φ(2 x e) Therefore,
φ(e) = φ(2 x e) / 2
Check:
φ(8) = φ(16) / 2
φ(8) = 4 ; φ(16) = 8, φ(16) / 2 = 4
Therefore,
φ(8) = φ(16) / 2
5. After working all the phi values above, and the tables I have realized that the φ value of a number n, where n is greater than 2, is always an even number.
φ(n > 2) = an even number

From the previous observations, it was obvious that the phi function should be investigated using prime numbers, as it was the only type of numbers that had a fixed formula. φ(p) = p – 1.
To make sure of my results ad deductions about the choice of prime numbers, I decided that I will create a table for the values of (n) and (n)²  and their factors beginning with n.

 (n) (n) ² φ(n) ² Factors 1 1 n.a. 2 4 2 2 x 1 3 9 6 3 x 2 4 16 8 4 x 2 5 25 20 5 x 4 6 36 12 6 x 2 7 49 42 7 x 6 8 64 32 8 x 4 9 81 54 9 x 6 10 100 40 10 x 4

From the above table it is seen that only prime numbers have a definite formula, and a relationship between all the numbers. For this reason I will continue my investigation using prime numbers.

In my next part of the investigation, I will investigate the phi values for the square of prime numbers. φ(p)²

Since the calculations of the phi values are too big in this part and the next parts of the investigations, I will use a calculator, which has a special function of finding the phi values. So for the phi values which are greater than 30 I will be using this calculator as the numbers are far too big to be calculated manually and will waste huge amounts of time.

 (p) (p)² φ(p)² Factors 2 4 2 2 x 1 3 9 6 3 x 2 5 25 20 5 x 4 7 49 42 7 x 6

Conclusion

which is equivalent to p x p(p – 1)

Check:

φ(p)³ = p(p – 1);

φ(2)³ =  2 x 2(2 – 1)

φ(8) = 4, 2 x 2(2 – 1) = 4

φ(2)³ = 2 x 2(2 - 1) (true)

φ(3)³ = 3 x 3(3 – 1)

φ(27) = 18, 3 x 3(3 – 1) = 18

φ(3)³ = 3 x 3(3 - 1) (true)

Now I investigate the value of the φ(p)^4. Just like the previous investigations I will use a table to help me.

 (p) (p) 4 φ(p)4 Factors 2 16 8 2 x 4 3 81 54 3 x 18 5 625 500 5 x 100

From the above table we can see that φ(p) 4 = p x φ(p)³ i.e.φ(p) 4 = p x p x p(p –1).

This is equivalent to φ(p) 4 = p4 – p³.

This can also be proved logically, as the prime number to the power of four, will be only divisible by the multiples of the prime number itself. the φ value of the prime number to the power of four will be equal to the prime number to the power of four minus the prime number to the power of 4 divided by the prime number itself. That is p^4 minus p^3. So this proves the formula φ(p) 4 = p4– p³, which is p x p x p (p – 1)

Check:

φ(p) 4 = p x p x p(p – 1);

φ(2) 4 =2 x 2 x  2(2 – 1)

φ(16) = 8, 2 x 2 x 2(2 – 1) = 8

φ(2) 4 = 2 x 2 x 2(2 - 1) (true)

φ(3) 4 = 3 x 3 x 3(3 – 1)

φ(81) = 54, 3 x 3 x 3(3 – 1) = 54

φ(3) 4= 3 x 3 x 3(3 - 1) (true)

We can say that:

φ(7) 4 = 7 x 7 x 7(7 – 1)

φ(2401) = 343(7 – 1) = 2058

Through investigating the φ(p), φ(p)², φ(p)³ and φ(p)4 I have found the following observations:

φ(p)=  (p – 1)

φ(p)²   =  p (p – 1)

φ(p)³   =  p² (p – 1)

φ(p) 4  = p³ (p – 1)

So we can actually say that:

φ(p) n  = p n – 1  ( φ(p) )

φ(p) n  = p n – 1  (p – 1)

This student written piece of work is one of many that can be found in our GCSE Phi Function section.

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