Can you explain again why there are no solutions to phi(theta)= 3 ? you said because phi(x) is even so only odd number allowed is 1. Is that a theorem or ?
Theta is a natural number, so it can be written as product of powers of prime. If p is prime divisor of theta, then p-1 divides phi(theta). But p-1 is even for any prime except 2, therefore, theta cannot have any prime factors other than 2. Now if theta is power of 2, say 2^k, then phi(2^k)=2^{k-1}. So, k cannot be greater than 1 to have an phi(theta)= odd number. However, if k=1, then you have phi(theta)=1. Therefore, the result.
One of the consequence of definition of phi function is phi(mn)=phi(m)phi(n) and phi(p^k)=p^k-p^(k-1). Now, phi(24)=phi(3x8)=phi(3)×phi(8)=2×4=8. Alternatively, since 24 is a "small" number, find all the numbers less than 24 and co-prime to 24. There are exactly 8 of those: 1, 5, 7, 11, 13, 17, 19, 23.
Great explanation ❤
Really helpful video.
Thank you sir.
Thank you
thank you sir
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hi! thanks for the video! I want to ask what if (p_i) | phi(n). what will be the consequences?
if p divides phi(n), and p>2, then p^k divides n for some k>1.
Can you explain again why there are no solutions to phi(theta)= 3 ? you said because phi(x) is even so only odd number allowed is 1. Is that a theorem or ?
Theta is a natural number, so it can be written as product of powers of prime. If p is prime divisor of theta, then p-1 divides phi(theta). But p-1 is even for any prime except 2, therefore, theta cannot have any prime factors other than 2.
Now if theta is power of 2, say 2^k, then phi(2^k)=2^{k-1}. So, k cannot be greater than 1 to have an phi(theta)= odd number. However, if k=1, then you have phi(theta)=1. Therefore, the result.
How find phi(24) ?
One of the consequence of definition of phi function is
phi(mn)=phi(m)phi(n) and phi(p^k)=p^k-p^(k-1).
Now, phi(24)=phi(3x8)=phi(3)×phi(8)=2×4=8.
Alternatively, since 24 is a "small" number, find all the numbers less than 24 and co-prime to 24. There are exactly 8 of those: 1, 5, 7, 11, 13, 17, 19, 23.