The Phi function.

This formula cannot be used to find out the Φ values for the number. It is just a relationship formula.

For example Φ(8) = 4. We multiply the result by 2 giving us 8. Now we check what 2n is. 2n is 16. So Φ(16) should be equal to 8 and Φ(16) is equal to 8.

We can take another number such as 10. The Φ(6) = 2 and then we multiply it by 2 and get 4. The Φ(12) should also be 4. This is true because Φ(12) = 4.

There is also another relationship formula.  It is Φ(10n) = 8[Φ(n)]

This we can prove

Φ(10 x 2) = 8 [Φ(2)]

Φ(20) = 8 x 1

Φ(20) = 8

This is true.

Φ(40) = 8[Φ(4)]

Φ(40) = 8 x 2

Φ(40) = 16

This too is true.

Thus we come to know that this above formula is true.

Next we will see if there is any relationship within the values for prime numbers.

From this we can see that the values of the phi of the prime numbers increases. The prime numbers have no other factors except for the original value and 1. So all the numbers before them are co-prime to them. There is no definite proportion as to their increase. But there is however a formula. We can see that the phi value for the number is one less then the number itself. For 11 the value is 10. For 2 the value is one. For 7 the value is 6 and so on. So the formula of obtaining the value of phi of a prime number will be            Φ(p) = p – 1. This formula fits as we have seen. p represents the prime number in the formula.

There is one observation that is common for all the different sets. This is the fact that the phi value for every number except for 2 is divisible by 2or in other words is a multiple of 2.

Now we will see which set of numbers can be used to investigate the phi function. The set of numbers above have to be chosen. Either it will be odd or even or prime so we have to choose from these three. The following table has been drawn up so that the chosen set of numbers is correct.

From the above table we can see that the prime numbers have a definite pattern. They have a constant formula as we have seen earlier. Therefore prime numbers are the most suitable to be used for this investigation. We will now see the investigation for the values of Φ(p)2. p stands for prime.

The following table lists all the prime numbers and their squares and the phi values of those squares.

We can deduce the formula. The formula will be Φ(p)2 = p(p-1). This we got because the Φ(p) = p-1 and so the p is multiplied to the formula and the so the above formula is obtained.

This formula has been explained logically and it can also be proved using sequences. The numbers that are taken are

(2,2) (3,6) (5,20) The number on the left is p and is regarded as x and the number on the right is the Φ(p2) for the corresponding value. This is substituted as y in the equation which is y = ax2 + bx +c. Thus there will be 3 equations once we substitute the 3 sets of numbers.

20 = 25a + 5b + c    ……..1

6= 9a + 3b + c  ……..2

2= 4a + 2b + c ……..3

We then subtract the second equation from the first one. The result will be

14 = 16a + 2b ……4

Then the second has to be subtracted from the third.

4 = 5a + b ………5

Now we can solve equation 4 and 5 simultaneously

14 = 16a + 2b

(4 = 5a + b) x –2

14 = 16a + 2b

-8 = -10a – 2b

6 = 6a

so a=1

Then we substitute a in the fourth equation

14 = 16 x 1 + 2b

-2 = 2b

b = -1

a=1     b = -1

Now the values for a and b have to substituted in the first equation

20=25 – 5 +c

20 = 20 +c

c=0

thus a=1, b = -1 and c=0 By entering these values into the equation y = ax2 + bx + c the final equation will be y = x2 –x . Thus the answer is

Φ(p)2 = p(p-1)

Now we will check this.

Φ(2)2 = 2(2-1)

Φ(4) = 2, 2(2-1) =2

So the above formula is correct.

So it can be said that

Φ(13)2 =13(13-1)

Φ(169) = 156

Therefore the formula has been proved.

Since we have to find out the formula for Φ(p)n, we also have to have to find out the values of Φ(p)3 . This we will do now.

The formula for this would be Φ(p)3 = p x Φ(p)2. This is the same as p x p(p-1). This is because p has been multiplied to p2 to give p3 and thus will also be multiplied to p(p-1) to give p x p(p-1). This formula can also be derived by using the formula y = ax3 + bx2 + cx + d because this is for cubic functions but it will not be used because it will be longer and more difficult to understand. Thus it would be better to understand it logically.

Now we will check this formula with some numbers.

Φ(p)3 = p x p(p-1)

Φ(2)3 = 4(1)

Φ(8) = 4

This is true.

For another value

Φ(5)3 = 25(4)

Φ(125) = 100

This is also true.

This proves that Φ(p)3 = p2(p-1)

Now I will investigate the same way for Φ(p)4

Taking bigger values for p would complicate the investigation and if the formula has to be found out using the sequence method, the formula now would be y = ax4 + bx3 + cx2 + dx + e. This again would be long and so we find out the formula logically. Multiplying p to p3 would give p4. Multiplying p2(p-1) by p would give p3(p-1) and this would be the formula. First we have to check it.

Φ(p)4 = p3(p-1)

Φ(2)4 = 8(1)

Φ(16)= 8

This is true.

Φ(11)4 = 1331(10)

Φ(14641) = 13310

This also is true. So we can understand that this formula is correct.

We have found out the following formulae

Φ(p) = p-1

Φ(p)2 = p(p-1)

Φ(p)3 = p2(p-1)

Φ(p)4 = p3(p-1)

This will continue as the power of p increase in the same pattern. Thus the formula for this would be

Φ(p)n = pn-1(p-1)

I will investigate Φ(mn) = Φ (m) x Φ(n)

I will check if this is possible every time with different number combinations. The different number combinations that I have used are:

1. Odd and Odd
2. Odd and Even
3. Odd and Prime
4. Prime and Prime
5. Prime and Even
6. Even and Even

These numbers will be substituted as m and n respectively. The phi of these numbers has already been seen in the table derived.

1. Odd and Odd

Odd numbers include 3, 5, 7, 9,11 and so on.

The phi values are listed below of some of the odd numbers.

Φ(3) = 2

Φ(5) = 4

Φ(7) = 6

Φ(9) = 6

Φ(11) = 10

Φ(13) = 12

Φ(15) = 8

Φ(17) = 16

These are the values of some of the phi values of odd numbers that we will consider. We have to check where m is an odd number and so is n.

Φ(3) x Φ(5) = Φ(15)

2 x 4 =8

Φ(15) = 8

Here the above formula is correct.

Φ(3) x Φ(9) = Φ(27)

2 x 6 =12

Φ(27) = 12

Here this does not apply.

Φ(5) x Φ(13) = Φ(65)

4 x 12 = 48

Φ(65) = 48

This apply here.

Φ(7) x Φ(3) = Φ(21)

6 x 2 = 12

Φ(21) = 12

Here the above formula applies.

Overall conclusion: The formula Φ(mn) = Φ(m) x Φ(n)  does not apply when both m and n are odd numbers.

2. Odd and Even

Φ(3) = 2                        Φ(2) = 1

Φ(5) = 4                        Φ(4) = 2

Φ(7) = 6                        Φ(6) = 2

Φ(9) = 6                        Φ(8) = 4

Φ(11) = 10                        Φ(12) = 4

Φ(13) = 12                        Φ(14) = 6

Φ(15) = 8                        Φ(16) = 8

Φ(17) = 16                        Φ(18) = 6

Above are the listed phi values for the odd and even numbers.

We have to check Φ(mn) = Φ(m) x Φ(n) where m is an odd number and n is an even number or vice versa.

Φ(11) x Φ(2) = Φ(22)

10 x 1 = 10

Φ(22) =10

This will apply here.

Φ(9) x Φ(4) = Φ(36)

6 x 2 =12

Φ (36) = 12

We can see that the formula does apply here.

Φ(7) x Φ (4) = Φ(28)

6 x 2 = 12

Φ(28) =12

This is true.

Φ(15) x Φ(6) = Φ(90)

8 x 2 = 16

Φ(90) = 16

This is false.

Overall conclusion: We can see that the above formula Φ(mn) = Φ(m) x Φ(n) is true except when m and n both are multiples of 5 nor are they both multiples of 3. This is true for all the odd numbers which when multiplied give an even number.

Check

Φ(5) x Φ(12) = Φ(60)

4 x 4 = 16

Φ(60) =16

Φ(3) x Φ(6) = Φ(18)

2 x 2 = 4

Φ(18) = 6

This is false because 3 and 6 are multiples of 3.

Thus the above statement is proved.

3. Odd and Prime

This where one number is odd and the other is prime.

Φ(3) = 2                        Φ(2) = 1

Φ(5) = 4                        Φ(3) = 2

Φ(7) = 6                        Φ(5) = 4

Φ(9) = 6                         Φ(7) = 6

Φ(11) = 10                        Φ(11) = 10

Φ(13) = 12                        Φ(13) = 12

Φ(15) = 8                        Φ(17) = 16

Φ(17) = 16                        Φ(19) = 18

Φ(3) x Φ(3) = Φ(9)

2 x 2 = 4

Φ(9) = 4

This is false.

Φ(9) x Φ(2) = Φ(18)

6 x 1 = 6

Φ(18) = 6

This is true.

Φ(3) x Φ(9) = Φ(27)

2 x 6 = 12

Φ(27) = 12

This is false.

Φ(2) x Φ(15) = Φ(30)

2 x 8 = 16

Φ(30) = 16

This is also true.

Overall conclusion: The formula Φ(mn) = Φ (m) x Φ (n) where one is odd and the other is prime is true for all conditions except when the same number is repeated as odd and prime such as 3, 5,7 etc. It is also not true when the phi of the number 3 is multiplied by the phi of its multiple. The same applies for 5 and its odd multiples.

Check

Φ(5) x Φ(9) = Φ(45)

4 x 6 = 24

Φ(45) = 24

This is true

Φ(5) x Φ(15) = Φ(75)

4 x 8 = 32

Φ(75) = 32

This is false.

Φ(3) x Φ(15) = Φ(45)

2 x 8 = 16

Φ(45) = 16

This is false.

Φ(9) x Φ(7) = Φ(63)

6 x 6 = 36

Φ(63) = 36

This is also true

Thus the above statement is proved.

4. Prime and Prime

Here I will investigate the formula Φ(mn) = Φ(m) x Φ(n) where both numbers are prime.

Φ(2) = 1

Φ(3) = 2

Φ(5) = 4

Φ(7) = 6

Φ(11) = 10

Φ(13) = 12

Φ(17) = 16

Φ(19) = 18

Φ(2) x Φ(3) = Φ(6)

1 x 2 = 2

Φ(6) = 2

This is true.

Φ(5) x Φ(5) = Φ(25)

4 x 4 = 16

Φ(25) = 16

This is false here.

Φ(3) x Φ(3) = Φ(9)

2 x 2 = 4

Φ(9) = 4

Thus it is false here again.

Φ(7) x Φ(5) = Φ(35)

6 x 4 = 24

Φ(35) =24

This is true.

Overall conclusion: If we look at the above numbers we will notice that there formula is true only when m and n are different prime numbers. To further check it we will take three numbers.

Φ(2) x Φ(3) x Φ(5) = Φ(30)

1 x 2 x 4 = 8

Φ(30) = 8

This is true.

The same we will take one which is repeated:

Φ(2) x Φ(2) x Φ(3) = Φ(12)

1 x 1 x 2 = 2

Φ(12) = 2

This is false, proving that even if the numbers are increased, the prime number should not be repeated. So for example Φ(24), Φ(36) cannot be found out using this method because they have the same prime factors repeating.

5. Prime and even

Here we will check if Φ(mn) = Φ(m) x Φ(n) is true when m and n are prime and even.

Φ(2) x Φ(2) = Φ(4)

1 x 1 = 1

Φ(4) =1

This is not true.

Φ(3) x Φ(8) = Φ(24)

2 x 4 = 8

Φ(24) = 8

This is true.

Φ(5) x Φ(6) = Φ(30)

4 x 2 = 12

Φ(30) = 12

This is false.

Φ(4) x Φ(2) = Φ(8)

2 x 1 = 2

Φ(8) = 2

Thus this too is also false.

Overall Conclusion: The above formula is true only when 3 is used as one number and a number other than a multiple of three is used. For example

Φ(3) x  Φ(6) = Φ(18)

2 x 2 = 4

Φ(18) = 4

This is false.

Φ(3) x Φ(10) = Φ(30)

2 x 4 = 8

Φ(30) = 8

This is true.

Φ(3) x Φ(8) = Φ(24)

2 x 4 = 8

Φ(24) = 8

This is also is true.

The above conclusion has been proved.

6. Even and even

Here we will see when m and n in the formula are substituted by even numbers.

Φ(4) x Φ(4) = Φ(16)

2 x 2 = 8

Φ(16) = 8

This is false.

Φ(6) x Φ(4) = Φ(24)

2 x 2 = 4

Φ(24) = 8

This again is false.

Φ(8) x Φ(4) = Φ(32)

4 x 2 = 8

Φ(32) = 8

This is false.

Φ(2) x Φ(4) = Φ(8)

1 x 2 =2

Φ(8) = 2

This is false.

Overall conclusion: From the above pattern we can see that although there is no direct relation as such, we can understand that if 2 is multiplied to most of the numbers it will give the correct phi. So the formula for this is actually Φ(mn) = Φ(m) x Φ(n) x 2

For example

2 x Φ(4) x Φ(6) = Φ(24)

2 x 2 x 2 = 8

Φ(24) = 8

This is true.

2 x Φ(8) x Φ(2) = Φ(16)

2 x 4 x 1 = 8

Φ(16) =8

This is also true.

So we can see that the formula Φ(mn) = Φ(m) x Φ(n) x 2 is true only when m and n both are even.