@@MyOneFiftiethOfADollar No, I'm an old engineer/programmer who enjoys math a lot, in particular NumberTheory, Algebraic Numbers, and Abstract Algebra. Keep up your good work!
Why doesn't it work for 4? Because 4 followes the second case and in there 4 = 2² and we relied on the fact that there exists two factors in the factorial j and j²-j and with 4 those two are no longer two anymore, they're just one number j = 2 j²-j = 2²-2 = 4-2 = 2. So that's why it doesn't work with 4. Is there any other number where this happens? Well, it only happens when j and j²-j equal one thing. j = j²-j Remove j from both sides j-j = j²-j-j 0 = j²-2j j²-2j = 0 Factor a j out j(j-2) = 0 Either j = 0, which is always false, or j-2 = 0, which is only true when j = 2. Remember, n = j², so this only happens with n = 2² = 4. 2j = j²
The only even prime , 2, is frequently a pathology. You answered your own question nicely. Was mentioned briefly at beginning that 4 does not divide (4-1)!=6. Note that the first odd prime squared is “big enough” to fit second case where 9 divides 8!= 2x3x4x5x6x7x8 since two multiples of 3 are present, 3 and 6 The second case was only case that made the problem non-trivial
Very clear proof for both cases, thanks!
Glad it made sense to you. Are you a number theory student?
@@MyOneFiftiethOfADollar No, I'm an old engineer/programmer who enjoys math a lot, in particular NumberTheory, Algebraic Numbers, and Abstract Algebra. Keep up your good work!
Do algebraic numbers have many applications in engineering/ programming?
They are zeroes of a polynomial right(not transcendental?)
Why doesn't it work for 4? Because 4 followes the second case and in there 4 = 2² and we relied on the fact that there exists two factors in the factorial j and j²-j and with 4 those two are no longer two anymore, they're just one number
j = 2
j²-j = 2²-2 = 4-2 = 2.
So that's why it doesn't work with 4.
Is there any other number where this happens?
Well, it only happens when j and j²-j equal one thing.
j = j²-j
Remove j from both sides
j-j = j²-j-j
0 = j²-2j
j²-2j = 0
Factor a j out
j(j-2) = 0
Either j = 0, which is always false, or j-2 = 0, which is only true when j = 2.
Remember, n = j², so this only happens with n = 2² = 4.
2j = j²
The only even prime , 2, is frequently a pathology. You answered your own question nicely.
Was mentioned briefly at beginning that 4 does not divide (4-1)!=6.
Note that the first odd prime squared is “big enough” to fit second case where 9 divides 8!= 2x3x4x5x6x7x8 since two multiples of 3 are present, 3 and 6
The second case was only case that made the problem non-trivial