Congrats on 50k subscribers man, you really deserve them ! Just one thing, the audio is hit or miss depending on the videos, perhaps you should change your gear if you can or master the sound manually. Cheers !
I think I found another way of solving this without using Catalan's conjecture. As 3 can't divide both k-1 and k+1 we can consider two cases: 1) k-1=2^x * 3^b-2 and k+1=2^a-x so 2^x * 3^b-2 + 1 = 2^a-x - 1 2^x-1 * 3^b-2 + 1 = 2^a-x-1, left side must be divisible by 2 so x=1; the case 2^a-x-1 = 1 has obviously no solutions for b 3^b-2 + 1 = 2^a-2, as 3^b-2 + 1 is not congruent to 0 mod 8 for all b we have to cases a=3 => b=2 => c=9 a=4 => b=3 => c=21 2) k+1=2^x * 3^b-2 and k-1=2^a-x so 2^x * 3^b-2 - 1 = 2^a-x + 1 2^x-1 * 3^b-2 = 2^a-x-1 + 1, left side can't be divisible by 2 so x=1; the case 2^a-x-1 = 1 gives x=2, a=2 and b=2 so c is not an integer 3^b-2 = 2^a-2 + 1 solving for b=2 and b=3 is trivial consider 2^x + 1 mod 27, we see that x is congruent to 9 mod 18 which is easy to check as 27 divides 2^9 + 1 (2^9 is congruent to -1 so 2^18 is congruent to 1 mod 27) but x is congruent to 9 mod 18 so 19 divides 2^x + 1 which is again easy to check (2^9 is congruent to -1 so 2^18 is congruent to 1 mod 19) so equation 3^b-2 = 2^a-2 + 1 has no solutions for b > 3. Now it's sufficient to check the cases a=0 and b=0.
For the a b > 0 solution I put a = p+q and b = u+v so the simultaneous equations give: 2^p.3^u + 2^q.3^v = 2c 2^p.3^u - 2^q.3^v = 6 From the second we can divide through by 6 and factorise to get 2^(q-1).3^(v-1) = 1 and 2^(p-q).3^(u-v) - 1 = 1 This gives us p=2 and q=u=v=1. So a=3, b=2. From the first equation we get c= 9.
my attempt before watching: 9(2^a*3^b-2+1) = c^2 so c=3k, where k is integer 9(2^a*3^b-2+1) = 9k^2 2^a*3^b-2+1 = k^2 modulo 6: k^2 is congruent to 1, so k=6n+1, where n is integer substitute in c=3k, and as a result c=18k+3 Checking some integers gives one solution: a = 4,b = 3,c = 21. Upd: wow, I missed so many sols, even with a and b being equal to zero
3^x - 3^y = 6 implies that 3^(x - 1) - 3^(y - 1) = 2, and that they are both integers. But then one of 3^(x - 1), 3^(y - 1) = 1, since otherwise 2 would be divisible by 3, and it’s immediate that we must have 3^(y - 1) = 1. Then 3^(x - 1) = 3 follows immediately. I forgot what the other case you asked about was but I imagine it proceeds similarly.
Ergh... Are we supposed to know and use Catalan's conjecture which has been solved 20 years ago? Not saying your proof is wrong, but it sounds like a Deus Ex Machina.
Yeah, it's not great. When I reached the fact the difference of the powers must be equal to 1, I found the three solution but had no way to prove they were the only ones. Frustrating.
Yeah there's got to be another way, especially considering this was a junior mathematical Olympiad problem Edit: on the Wikipedia page for Catalans conjecture, it is mentioned that the special case that is relevant in the video was first proved in the 1300's so presumably there is an elementary solution.
@@akaRicoSanchez these should give you rigorous proof : case 2 take %4 [which includes case 1] & case 3 take %4 and then %3 ....each has 2-3 cases looking at parity
You are free to suggest your own methods, why aren't there any? As for myself, I am glad that I have learned a theorem, which I haven't heard about before
@@PerfectCourtier After checking for the case of a-2=b-2=1, you can assume they are both bigger than one. Then you can look at decompositions of 3^(b-2)-1 and 2^(a-2)-1 modulo 2 and 3 respectively to determine that b-2 and a-2 have to be even in each of the respective cases. Then use difference of squares and you're done.
Congrats on 50k subscribers man, you really deserve them !
Just one thing, the audio is hit or miss depending on the videos, perhaps you should change your gear if you can or master the sound manually.
Cheers !
So true
I think I found another way of solving this without using Catalan's conjecture.
As 3 can't divide both k-1 and k+1 we can consider two cases:
1) k-1=2^x * 3^b-2 and k+1=2^a-x
so 2^x * 3^b-2 + 1 = 2^a-x - 1
2^x-1 * 3^b-2 + 1 = 2^a-x-1, left side must be divisible by 2 so x=1; the case 2^a-x-1 = 1 has obviously no solutions for b
3^b-2 + 1 = 2^a-2, as 3^b-2 + 1 is not congruent to 0 mod 8 for all b we have to cases
a=3 => b=2 => c=9
a=4 => b=3 => c=21
2) k+1=2^x * 3^b-2 and k-1=2^a-x
so 2^x * 3^b-2 - 1 = 2^a-x + 1
2^x-1 * 3^b-2 = 2^a-x-1 + 1, left side can't be divisible by 2 so x=1; the case 2^a-x-1 = 1 gives x=2, a=2 and b=2 so c is not an integer
3^b-2 = 2^a-2 + 1
solving for b=2 and b=3 is trivial
consider 2^x + 1 mod 27, we see that x is congruent to 9 mod 18 which is easy to check as 27 divides 2^9 + 1 (2^9 is congruent to -1 so 2^18 is congruent to 1 mod 27)
but x is congruent to 9 mod 18 so 19 divides 2^x + 1 which is again easy to check (2^9 is congruent to -1 so 2^18 is congruent to 1 mod 19)
so equation 3^b-2 = 2^a-2 + 1 has no solutions for b > 3.
Now it's sufficient to check the cases a=0 and b=0.
For the a b > 0 solution I put a = p+q and b = u+v so the simultaneous equations give:
2^p.3^u + 2^q.3^v = 2c
2^p.3^u - 2^q.3^v = 6
From the second we can divide through by 6 and factorise to get
2^(q-1).3^(v-1) = 1 and
2^(p-q).3^(u-v) - 1 = 1
This gives us p=2 and q=u=v=1.
So a=3, b=2.
From the first equation we get c= 9.
Great solution 👍🏼👍🏼
my attempt before watching:
9(2^a*3^b-2+1) = c^2
so c=3k, where k is integer
9(2^a*3^b-2+1) = 9k^2
2^a*3^b-2+1 = k^2
modulo 6: k^2 is congruent to 1, so k=6n+1, where n is integer
substitute in c=3k, and as a result c=18k+3
Checking some integers gives one solution: a = 4,b = 3,c = 21.
Upd: wow, I missed so many sols, even with a and b being equal to zero
What's the reasoning behind the only solutions to 3^x - 3^y = 6 and the same for 2^. What's the proof that the given solutions are the only ones.
3^x - 3^y = 6 implies that 3^(x - 1) - 3^(y - 1) = 2, and that they are both integers. But then one of 3^(x - 1), 3^(y - 1) = 1, since otherwise 2 would be divisible by 3, and it’s immediate that we must have 3^(y - 1) = 1. Then 3^(x - 1) = 3 follows immediately. I forgot what the other case you asked about was but I imagine it proceeds similarly.
Ergh... Are we supposed to know and use Catalan's conjecture which has been solved 20 years ago? Not saying your proof is wrong, but it sounds like a Deus Ex Machina.
Yeah, it's not great. When I reached the fact the difference of the powers must be equal to 1, I found the three solution but had no way to prove they were the only ones. Frustrating.
Yeah there's got to be another way, especially considering this was a junior mathematical Olympiad problem
Edit: on the Wikipedia page for Catalans conjecture, it is mentioned that the special case that is relevant in the video was first proved in the 1300's so presumably there is an elementary solution.
@@akaRicoSanchez these should give you rigorous proof : case 2 take %4 [which includes case 1] & case 3 take %4 and then %3 ....each has 2-3 cases looking at parity
You are free to suggest your own methods, why aren't there any?
As for myself, I am glad that I have learned a theorem, which I haven't heard about before
@@PerfectCourtier After checking for the case of a-2=b-2=1, you can assume they are both bigger than one. Then you can look at decompositions of 3^(b-2)-1 and 2^(a-2)-1 modulo 2 and 3 respectively to determine that b-2 and a-2 have to be even in each of the respective cases. Then use difference of squares and you're done.
A=2 B=1 C=4
Çok güzel 👍👏
3,2,9 is also a solutions
solved at 9:22
find sum of all integers n ≥ 1 such that 2n−1 has exactly n positive integer divisors
Hi you can try this if u want (:
JBMO
nice