Does this not fit using mathematical induction? Might be exactly on that way. I mean proving the left with the right side. I got the way but ... seems not true. No offense. :)
Let z1 and z2 be the roots if quadratic eq. z1 +z2 = -b/a (1) z1 . z2 = c/a (2) By taking mod on both sides of eq 1 , we will get angle between z1 & z2 120 degrees. Now we can write z2 = z1 e^i(120) Substitute z2 in eq 1 and 2 and square eq 1 to get the desired result
Alternatively: |a|=|b|=|c|=|Z|=1 so conjugate of a =1/a same for b,c and z( I hope you know for a complex no a , a.ā=|a|²). Now take conjugate of eqn az²+bz+c=0. substitute the conjugates of a,b,c,z by their reciprocal (as obtained earlier) you get (abz²+acz+bc=0). so Z satisfies az²+bz+c=0...(1) abz²+acz+bc=0.....(2) Now do bc×(1)-c×(2) you get (b²-ac)Z=0 but since given that Z is non zero you can conclude b²=ac
@@arpangoswami5760 you can also just use quadratic equation to.solve for z in terms of and b and then plug this in to the b squared equation..see what I mean?
That explaination in the video are skipped some logic .Let me do some step by step. 1. n! divisible by n ( easy) 2. let n be a even number n= 2*m 3 n! divisible by 2*m 4 n! divisible by 2*m*n . this step may be a bit difficult . Remember that n>m and >2 . n! at some point need to have m and n and 2 in it . let say n= 8 . n! = 8*7*6*5*4*3*2 . all 2mn are there. 5. 2*m*n= n^2 6 We proof that n! divisible by n^2 . give condition that n is even number. >= 6. this is itself a good question.
@@bosorot Thank you, although I understood all how it works. My concern was more how one comes up with such a nice argument. Probably this is what makes one a good mathematician.
all of the questions can be solved by just throwing in some dummy numbers, like -1, 0, 1, 2, 3, but I just do not know how to prove that these are the only solutions :(
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Actually, n!≡-1(mod (n+1))
So by Wilson’s Theorem
Yes,(n+1) is prime
You mean converse of Wilson's Theorem
What if you don't know that tjeorem..can't younsolve without mod..
Does this not fit using mathematical induction? Might be exactly on that way. I mean proving the left with the right side. I got the way but ... seems not true. No offense. :)
Wait just because p prime divided n plus 1 does not tell you for certain it divides n! plus 1...
In case anyone else stumbles upon this, the question implies it
Hi can you solve this problem. Suppose a,b,c,z are nonzero complex numbers such that |a|=|b|=|c|, |z|=1 and az^2+bz+c=0. Show that b^2=ac.
Let z1 and z2 be the roots if quadratic eq.
z1 +z2 = -b/a (1)
z1 . z2 = c/a (2)
By taking mod on both sides of eq 1 , we will get angle between z1 & z2 120 degrees.
Now we can write z2 = z1 e^i(120)
Substitute z2 in eq 1 and 2 and square eq 1 to get the desired result
thanks for the help :)
👍
Alternatively: |a|=|b|=|c|=|Z|=1 so
conjugate of a =1/a
same for b,c and z( I hope you know for a complex no a , a.ā=|a|²). Now take conjugate of eqn az²+bz+c=0. substitute the conjugates of a,b,c,z by their reciprocal (as obtained earlier)
you get (abz²+acz+bc=0). so Z satisfies
az²+bz+c=0...(1)
abz²+acz+bc=0.....(2)
Now do bc×(1)-c×(2)
you get (b²-ac)Z=0 but since given that Z is non zero you can conclude b²=ac
@@arpangoswami5760 you can also just use quadratic equation to.solve for z in terms of and b and then plug this in to the b squared equation..see what I mean?
Nice
n² = 2mn
n² | n! ?
becasue 2 and m are smaller than n and m is not equal to 2.
2, m and n are all factors different from each other of n! therefore 2mn divides n!
@@RelaxJigglypuff thank you very much
I understand now
n>m>2>0
n² = 2mn | n! = n(n-1)...3•2•1
m = n-1, n-2, ... , 4 or 3
@@lucasmartiniano6915 thank you very much
I understand now
n>m>2>0
n² = 2mn | n! = n(n-1)...3•2•1
m = n-1, n-2, ... , 4 or 3
Can't we just use converse Wilson s theorem to get n+1 is prime?
Nice, how did you get to considering n^2 ?
That explaination in the video are skipped some logic .Let me do some step by step.
1. n! divisible by n ( easy)
2. let n be a even number n= 2*m
3 n! divisible by 2*m
4 n! divisible by 2*m*n . this step may be a bit difficult . Remember that n>m and >2 . n! at some point need to have m and n and 2 in it . let say n= 8 . n! = 8*7*6*5*4*3*2 . all 2mn are there.
5. 2*m*n= n^2
6 We proof that n! divisible by n^2 . give condition that n is even number. >= 6. this is itself a good question.
@@bosorot Thank you, although I understood all how it works. My concern was more how one comes up with such a nice argument. Probably this is what makes one a good mathematician.
@@tomask7732
I guess, you could say brute-forcing? But then, experience is the best.
i did it different way, that probably is easier to invent- but also it is much longer solution than yours, therefore i like yours more :)
Can you share your method I'm curious?
Thanks a Zettabyte ❤🌿🌿
JEE ADVANCE
@@mustafizrahman2822 Yeah!
@@jimmykitty 😁
@@mustafizrahman2822 Hey! How's your study going?
Excellent!
Nice
asnwer= 1
Also 2 and 4...
all of the questions can be solved by just throwing in some dummy numbers, like -1, 0, 1, 2, 3, but I just do not know how to prove that these are the only solutions :(
1