2017^a odd => b⁶-32b+1 odd => b even Setting b=2m, 2017^a = 64(b⁶-b)+1. 2017 = 33 mod 64, so 33^a = 1 mod 64 => a is even (note that (2n+1)² = 4n²+4n+1 = 1 mod 4n, so in particular, 33²=1 mod 64). So set a=2n, and factorize: 2017^2n = b⁶-32b+1 => (b³)²-(2017^n)²=32b-1 But LHS factorizes as (b³+2017^n)(b³-2017^n). So either LHS=RHS=0 => 32b=1 (not possible) or |b³-2017^n| ≠ 0 => |32b-1| ≥ |b³+2017^n| => 32b > b³ => b² < 32 => b < 6. This just leaves b = 2 and b = 4 to check manually. For b=2, RHS=64-64+1=1, which gives a=0 which is not in IN. For b=4, RHS=4096-128+1, clearly between a=1 and a=2. So no solutions exist.
LHS is odd, so b has to be even. Therefore 2017^a-1 is a multiple of 64. Since 2017-1=2016=32*63, this can only happen when a is even. So we can rule out the case of a being odd immediately.
@@maximilianveyne3973 it misses a few steps, but 2017^a - 1 = 2017^a - 1^a which can be factored into 2016 times sum of terms in form 2017^(a-1-n), where n goes from 0 to a-1. For odd a, you will have odd number of powers of 2017 summed, which gives an odd number, so it doesn't have any 2 as a factor.
First we see that b is odd so b can be represented as 2k+1. If we take everything mod 4 then we see that there is no solution for all cases. 2017^a mod 4 = 1^a=1, but (2k+1)^6 - 32*(2k+1) + 1 = 6*2k + 1 + 1 = (we use Pascal's triangle to help expand the term (2k+1) and use mod 4 to get rid of all terms where (2k)^n is greater than 1 since then (2k)^n = 0 mod 4) = 2 mod 4.
a=odd case can be proved as follows. Write the equation as 2016*(2017^(a-1)+2017^(a-2)+...+1) = b^6 - 32b. Obviously b must be even which means 64 divides RHS. But 64 does not divide LHS if a is odd, since 2016=32*63 and the second factor is odd. I think it is the intended solution.
Isn't there much easier to see that b has to be even (2017^a - 1 - 32b is even) so we have: b = 2*c 2^6(c^6-c) = -1 mod 2017 = 2016 mod 2017 But 2016 = 2^5 *63 so 2*c*(c^5-1) = 63 mod 2017 not possible (left side is even, right is odd) Unless we allow zeros of course.
@@spiderjerusalem4009 As far as I know, that depends on the area of mathematics. In number theory, zero usually is not part of N; in real analysis, when talking about series etc., zero often is part of N.
@@wesleydeng71 In Set theory, 0 is a natural number, the first of them, represented by the empty set. It is also a Peano number, which also is another representation of natural numbers.
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In Hungary we also don't consider zero part of N But part of Z
Im curious, into what, idk how to describe its like hmm, the successor function must iterate before multiplication is valid, uncountable infinity while equal and only cardinality greater than inf, there are methods that get u there quicker ... i dont even know what im trying to describe, obviously a successor is a part of logic, and given a proposition there are true entailments and false entailments, and those all exist instantaneously But it feels like, if we were to bootstrap or compile said logic it would fail in a way, that while sure "yes multiplication exists" that only exists on the bootstrap of the previous "addition compiler" Im sorry my ask is so vague, i dont know how else to explain what im asking 😪, maybe im asking what satisfiability critereon need for a higher order logic? But there's also formulations where u can show that it grows uncountably infinite quicker by increasing N which just isnt... idk just infinity*infinity I feel like math is missing something as stupid as 0.9 repeating + epsilon is 1, not merely == 1, to say that limit as 5x -> inf, vs 3x -> inf, 5x > 3x as its true at every evaluation point i understand why all the above are true, and nothing is contradiction in accepted style, but i feel like theyre all pointing at an object, or a style of math ... what is that math?
Since 0 for a has been excluded at the beginning, the result is kind of disappointing. Including 0 as possible values for a and b, there would be 2 solutions at the end.
Oh I forgot n logbase[x^n](1+a) = logbase[x^n](1^n + a^n) = logbase[x^n]((1+a)^n) Divide both sides by n 1/n + logbase[x^n](1+a) - logbase[x^n](a) - logbase[x^n](b) - 1 = 0 If the math is correct 🧐🤪
As he stated multiple times, 0 is not being considered a natural number for the problem and therefore a and b cannot equal 0. He said it is a solution in terms of integers, just not naturals.
I think I have found a much shorter proof, but I'm unsure whether it's sketchy somewhere (edit: it is): - First, lets look at the equation where we rearrange the 1 to the left hand side and reduce mod 2016: the left hand side is now 2017^a - 1, and since 2017 is congruent to 1 mod 2016, the left hand side must be congruent to 0 mod 2016 for all a positive integers. - This implies the (rearranged) right hand side must also be congruent to 0 mod 2016, therefore: b^6-32b == 0 mod 2016, b^6 == 32b mod 2016, b^6 == 2^5*b mod 2016, 2^5 == b^5 mod 2016 - Now if b^5 is congruent to 32 mod 2016, then b^5 == 31 mod 2017. In the original equation, the left hand side is congruent to 0 mod 2017, so we look at the right hand side: 31*b-32*b+1 == -b+1 == 0 mod 2017, b == 1 mod 2017. Since b cannot simultaneously be congruent to 1 and 31 mod 2017, we've reached a contradiction, and there is NO solution when a and b are both positive integers Edit: I initially assumed b^5==2^5 implied b==2, but thanks to feedback by @egoreremeev9969 I've changed that and hope the proof is correct now.
if a^b == c^b (mod d) it doesn't follow that a == c (mod d) Look at the squares example in the video. In mod 5 you get: n : 0, 1, 2, 3, 4 n^2: 0, 1, 4, 1, 4 This means that squaring is not reversible, so even though 2^2 == 4^2 (mod 5) it doesn't mean that 2 == 4 (mod 5)
@@egoreremeev9969 Thanks a lot for pointing that out! I remembered the opposite implication so that if a == b mod n then a^m == b^m mod n, and for some reason thought it worked both ways, but it makes all sense that it doesnt. Edit: Well, now that i think about it, even if we can't reverse the exponentiation, I think we could substitute b^5 == 32 mod 2016, which implies b^5 == 31 mod 2017 in the last step, getting: 31*b-32*b+1 == -b+1 == 0 mod 2017, b == 1 mod 2017. This leads to a contradiction still, since b == 1 mod 2017 means b^5 can't possibly be 31 mod 2017
Since 2016 is not prime, you cannot cancel out b. To see this, just consider mod 10 (that is, the last digit). If your number ends e.g. in a 4, then half the number may end in either a 2 or a 7. Thus 2a=2b (mod 10) does not imply a=b (mod 10). Only modulo prime you can cancel arbitrsry non-zero factors. Edit: You also missed the possibility that b=0 (mod 2016).
@@terak93 if b^5 is congruent to 32 mod 2016, then this does not directly imply that b^5 is congruent to 31 mod 2017. To illustrate this, note that if x is congruent to 1 mod 2, then x does not have to be 0 mod 3, i.e., a multiple of 3.
I'm surprised you didn't take advantage of the fact that 2017 is congruent to one, modulo 7, 8, and 9. Together with the CRT, this tells you that b must be one of { 0, 2, 56, 72 }, modulo 126. Then, the fact that 3^6 < 2017 < 4^6 shows that b > 3, which means that the smallest possible value of b is 56. Since b must be even, b^6 - 32*b + 1 == 1 (mod 64). Since 2017 == 33 (mod 64), and 33 has order 2, modulo 64, a must be even, which means that b^6 - 32*b + 1 must be a perfect square. But b^6 - 32*b + 1 is only a perfect square for b
@@mathemagicalpi by definiton 0 is a natural number. If 0 was not a natural number then there would be not need to call then "natural" since name "positive integers" exists already
b^6 - 32b + 1 for b=3 == 634 (not -634), but also not a perfect square.
I agree.
2017^a odd => b⁶-32b+1 odd => b even
Setting b=2m, 2017^a = 64(b⁶-b)+1.
2017 = 33 mod 64, so 33^a = 1 mod 64 => a is even (note that (2n+1)² = 4n²+4n+1 = 1 mod 4n, so in particular, 33²=1 mod 64).
So set a=2n, and factorize:
2017^2n = b⁶-32b+1
=> (b³)²-(2017^n)²=32b-1
But LHS factorizes as (b³+2017^n)(b³-2017^n).
So either LHS=RHS=0 => 32b=1 (not possible) or |b³-2017^n| ≠ 0 => |32b-1| ≥ |b³+2017^n| => 32b > b³ => b² < 32 => b < 6.
This just leaves b = 2 and b = 4 to check manually. For b=2, RHS=64-64+1=1, which gives a=0 which is not in IN. For b=4, RHS=4096-128+1, clearly between a=1 and a=2.
So no solutions exist.
LHS is odd, so b has to be even. Therefore 2017^a-1 is a multiple of 64. Since 2017-1=2016=32*63, this can only happen when a is even. So we can rule out the case of a being odd immediately.
If you observe it mod 7 and assume b is co prime to 7 than you get b=0 mod 7 which means b must be of form 7k dont ask how this helps
I understand up to the final step. Why does a=1 failing to be a multiple of 64 rule out every other possible odd value of a?
@@maximilianveyne3973 it misses a few steps, but 2017^a - 1 = 2017^a - 1^a which can be factored into 2016 times sum of terms in form 2017^(a-1-n), where n goes from 0 to a-1.
For odd a, you will have odd number of powers of 2017 summed, which gives an odd number, so it doesn't have any 2 as a factor.
First we see that b is odd so b can be represented as 2k+1. If we take everything mod 4 then we see that there is no solution for all cases. 2017^a mod 4 = 1^a=1, but (2k+1)^6 - 32*(2k+1) + 1 = 6*2k + 1 + 1 = (we use Pascal's triangle to help expand the term (2k+1) and use mod 4 to get rid of all terms where (2k)^n is greater than 1 since then (2k)^n = 0 mod 4) = 2 mod 4.
Is it just me reading the title as "What a nice Dophamine equation" after the "A very satisfying solution" on the thumbnail?
a=odd case can be proved as follows. Write the equation as 2016*(2017^(a-1)+2017^(a-2)+...+1) = b^6 - 32b. Obviously b must be even which means 64 divides RHS. But 64 does not divide LHS if a is odd, since 2016=32*63 and the second factor is odd. I think it is the intended solution.
Isn't there much easier to see that b has to be even (2017^a - 1 - 32b is even) so we have:
b = 2*c
2^6(c^6-c) = -1 mod 2017 = 2016 mod 2017
But 2016 = 2^5 *63 so
2*c*(c^5-1) = 63 mod 2017 not possible (left side is even, right is odd)
Unless we allow zeros of course.
With b = 3 the expression b^6.... is +634
Sad that zero doesn't make it into N in America.
but it seldom does no matter in what country
@@spiderjerusalem4009 As far as I know, that depends on the area of mathematics. In number theory, zero usually is not part of N; in real analysis, when talking about series etc., zero often is part of N.
Zero is definitely not "natural". It was invented by the Indians quite late in the history.
@@wesleydeng71 In Set theory, 0 is a natural number, the first of them, represented by the empty set. It is also a Peano number, which also is another representation of natural numbers.
In Hungary we also don't consider zero part of N
But part of Z
Im curious, into what, idk how to describe its like hmm, the successor function must iterate before multiplication is valid, uncountable infinity while equal and only cardinality greater than inf, there are methods that get u there quicker
... i dont even know what im trying to describe, obviously a successor is a part of logic, and given a proposition there are true entailments and false entailments, and those all exist instantaneously
But it feels like, if we were to bootstrap or compile said logic it would fail in a way, that while sure "yes multiplication exists" that only exists on the bootstrap of the previous "addition compiler"
Im sorry my ask is so vague, i dont know how else to explain what im asking 😪, maybe im asking what satisfiability critereon need for a higher order logic? But there's also formulations where u can show that it grows uncountably infinite quicker by increasing N which just isnt... idk just infinity*infinity
I feel like math is missing something as stupid as 0.9 repeating + epsilon is 1, not merely == 1, to say that limit as 5x -> inf, vs 3x -> inf, 5x > 3x as its true at every evaluation point
i understand why all the above are true, and nothing is contradiction in accepted style, but i feel like theyre all pointing at an object, or a style of math ... what is that math?
I can't understand your question. Are you asking about Peano arithmetic, ordinal number, or nonstandard analysis?
Bro never gets old
In other words, 2 and 3 are both quadratic non-residues modulo 5.
So innovative methods
12:31
Since 0 for a has been excluded at the beginning, the result is kind of disappointing.
Including 0 as possible values for a and b, there would be 2 solutions at the end.
the Square beyond compare !
The b=3 statement seems wrong. It's impossible that is -634
Well the negative is wrong, it is 3^6-32*3+1 = 634. This is still not a power of 2017.
It almost looks like:
1+ logbase[x^n](1+a^n) = n( logbase[x^n](a) + logbase[x^n](b) + 1)
Oh I forgot n logbase[x^n](1+a) = logbase[x^n](1^n + a^n) = logbase[x^n]((1+a)^n)
Divide both sides by n
1/n + logbase[x^n](1+a) - logbase[x^n](a) - logbase[x^n](b) - 1 = 0
If the math is correct 🧐🤪
Also that kind of looks like two vectors being subtracted from a larger vector no?
I did the homework for a=0. In that case b must be either 1 or 2.
0 or 2
the audio is very weird, seems to be detached from the video or something, a bit uncomfortable to listen to
7:06
Error.
If a is even, there is a solution : a=0 and b=2.
b^6 - 32*b + 1 = 1 = 2017^a
As he stated multiple times, 0 is not being considered a natural number for the problem and therefore a and b cannot equal 0. He said it is a solution in terms of integers, just not naturals.
Nice
i come here to watch magic
I think I have found a much shorter proof, but I'm unsure whether it's sketchy somewhere (edit: it is):
- First, lets look at the equation where we rearrange the 1 to the left hand side and reduce mod 2016:
the left hand side is now 2017^a - 1, and since 2017 is congruent to 1 mod 2016, the left hand side must be congruent to 0 mod 2016 for all a positive integers.
- This implies the (rearranged) right hand side must also be congruent to 0 mod 2016, therefore:
b^6-32b == 0 mod 2016, b^6 == 32b mod 2016, b^6 == 2^5*b mod 2016, 2^5 == b^5 mod 2016
- Now if b^5 is congruent to 32 mod 2016, then b^5 == 31 mod 2017. In the original equation, the left hand side is congruent to 0 mod 2017, so we look at the right hand side:
31*b-32*b+1 == -b+1 == 0 mod 2017, b == 1 mod 2017.
Since b cannot simultaneously be congruent to 1 and 31 mod 2017, we've reached a contradiction, and there is NO solution when a and b are both positive integers
Edit: I initially assumed b^5==2^5 implied b==2, but thanks to feedback by @egoreremeev9969 I've changed that and hope the proof is correct now.
if a^b == c^b (mod d) it doesn't follow that a == c (mod d)
Look at the squares example in the video. In mod 5 you get:
n : 0, 1, 2, 3, 4
n^2: 0, 1, 4, 1, 4
This means that squaring is not reversible, so even though 2^2 == 4^2 (mod 5) it doesn't mean that 2 == 4 (mod 5)
@@egoreremeev9969
Thanks a lot for pointing that out!
I remembered the opposite implication so that if a == b mod n then a^m == b^m mod n, and for some reason thought it worked both ways, but it makes all sense that it doesnt.
Edit:
Well, now that i think about it, even if we can't reverse the exponentiation, I think we could substitute b^5 == 32 mod 2016, which implies b^5 == 31 mod 2017 in the last step, getting:
31*b-32*b+1 == -b+1 == 0 mod 2017, b == 1 mod 2017.
This leads to a contradiction still, since b == 1 mod 2017 means b^5 can't possibly be 31 mod 2017
Since 2016 is not prime, you cannot cancel out b. To see this, just consider mod 10 (that is, the last digit). If your number ends e.g. in a 4, then half the number may end in either a 2 or a 7. Thus 2a=2b (mod 10) does not imply a=b (mod 10). Only modulo prime you can cancel arbitrsry non-zero factors.
Edit: You also missed the possibility that b=0 (mod 2016).
@@__christopher__
Oh, so this means I cannot assume b^5 == 2^5 ?
@@terak93 if b^5 is congruent to 32 mod 2016, then this does not directly imply that b^5 is congruent to 31 mod 2017. To illustrate this, note that if x is congruent to 1 mod 2, then x does not have to be 0 mod 3, i.e., a multiple of 3.
I'm surprised you didn't take advantage of the fact that 2017 is congruent to one, modulo 7, 8, and 9. Together with the CRT, this tells you that b must be one of { 0, 2, 56, 72 }, modulo 126. Then, the fact that 3^6 < 2017 < 4^6 shows that b > 3, which means that the smallest possible value of b is 56. Since b must be even, b^6 - 32*b + 1 == 1 (mod 64). Since 2017 == 33 (mod 64), and 33 has order 2, modulo 64, a must be even, which means that b^6 - 32*b + 1 must be a perfect square. But b^6 - 32*b + 1 is only a perfect square for b
2^6-32*2+1 = 1^2 is also a perfect square.
@@周品宏-o7w Good catch. I meant b
❤❤
I have always hated number theory, and now it's still not an exception.
Even worse is combinatorics... 😅 It's just a grab bag of tricks all the way down. 😅
a=0, b=0
That depends whether or not 0 is a natural number or not
@@mathemagicalpiit is.
@@mathemagicalpi by definiton 0 is a natural number. If 0 was not a natural number then there would be not need to call then "natural" since name "positive integers" exists already
also a=0, b=2
@bartekabuz855 How? What's your definition of natural numbers because when I look it up, 0 is not.
3rd comment alert!! Me!!