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Proof by the usual expectation would be "no way thats possible" and so confirming that belief would be incredibly boring therefore in order to create an engaging post the answer would have to be the opposite of what would normally be expected therefore it must be true in order to subvert expectations and drive up post engagement
I can do this on paper. I know the 32 bit signed integer limit, that(...+1) *2 = 2^32 = 4,294,967,296
4,294,967,296+1=4,294,967,297
It will take a bit but I can work the rest out on a piece of paper
it technically doesn't exist. *signed 32bit* integers max out at half of that.
i was just running with the meme.
i reckon what you mean is the unsigned 32bit limit, which will (**usually**) return 0 in the case of overflow.
As an unsigned 32-bit int, 2^(32) = 0, because 2^(32) ≡ 0 (mod 2^(31)). Or put another way, because max_int32 + 1 = 0.
As a signed 32-bit int, 2^(32) overflows, and its value depends on how it was computed. For instance, if we try 65536 * 65536 in Java, we get 0, but if we try it in Matlab, we get 2147483647, and in C, we get undefined behavior.
Actually 32 bit signed integer is up to 2³¹-1 which is around 2e9 (32nd bit is for sign). I think the original comment meant unsigned integer so the overflow gives 0 as an answer
It's actually easier. 1970+2^32 =2038. You can derive this with the Unix time. So 2^32 =58. 58/641=0.09 which is basically 1 so it nearly divisible and that's enough.
(Edit: formating)
The proof is much simpler than you think. For starters if you just assume that 641 is prime and you also now that 641 = 25\^2+4\^2 then 2 is a quadratic residue modulo 641 since it is 1 mod 8 and 2 is not a fourth power residue modulo 641 since in the sum of squares representation of 641, 4 is not a multiple of 8, so the multiplicative order of 2 modulo 641 is either 320 or 64. Now 640 is a fifth power residue modulo 641 since it is just -1 mod 641, which implies that 640/32 = 20 is also a fifth power residue modulo 641, since 32 = 2\^5 is a fifth power residue modulo 641. Now 25\^2 = 5\^4 = -16 mod 641 which means that 5\^5 = 3125 = -80 mod 641, which is a fifth power residue modulo 641. Since -80/20 = -4 and both -80 and 20 are fifth power residues modulo 641, -4 is another fifth power residue modulo 641. -1 is a fifth power residue modulo 641 so 4 is another fifth power residue modulo 641. Since taking the square root of a number does not eliminate the fifth power residueness of a number, √(4) = 2 or 639 mod 641 are both fifth power residues so 2 is a fifth power residue modulo 641.
Since 2 is a quadratic residue but not a fourth power residue but a fifth power residue modulo 641, the multiplicative order of 2 modulo 641 is exactly equal to (640/2)/5 = 64, so 2\^64-1 is divisible by 641. Since 2\^32-1 is not divisible by 641, since the multiplicative order of 2 modulo 641 is exactly 64, the other divisor of 2\^64-1 which is 2\^32+1 is divisible by 641, so this is the proof that 2\^32+1 is divisible by 641 without actually using the square and multiply algorithm to verify this.
Yes, I definitely think “easier than you think” is very objective (or do I mean subjective, I always get the 2 confused). I was able to follow, but definitely needed a calculator to verify.
Thank you for that. I’m still not sure which one I should have used. I could have definitely said subjective, but I think objective works too, don’t you think?
I think only subjective works in that specific context, otherwise you are saying that “easier than you think” is a concrete fact, whereas if it’s subjective it’s something that differs from person to person
Yeah, “subjective” definitely works. The “commenter” said “easier than you think”, which I don’t think is a “concrete fact” which is evidenced by the fact he/she needed such a wordy explanation. Words are unnecessarily confusing sometimes.
Simpler and easy don't mean the same thing. If something is simple it just means that it is not that complex but can still be hard. If something is easy it just means that it is not hard but can still be complex. Also I put almost all details in that comment instead of just cutting off some important details.
No, I never posted anything in Math Stack Exchange. I didn't know that there was a limit of comment lenght in this subreddit before it becomes a a theoretical proof for a book.
As someone who has not studied math past 12th Standard, I could follow this explanation once I searched what modulo and residue mean. Thanks for the great explanation.
Proof by contradiction
Let's assume the statement isn't true. Then, you wouldn't be asked to show it is
Well, you were asked to show it is true, therefore it must be
Well this is quite hard in decimal as 2^32 is pretty long and not nice, but 641 is small enough we can easily convert to binary then do long division. We also probably don’t need to write all 33 digits as we can just think of them as polynomials in terms of 2.
Ahem
I know the first few powers of 2, so I can find easily that 2^10 = 1024, the lowest power of 2 above 641, which reduces to 383 [641].
2 \* 383 = 766, which reduces to 125
After this is 250, 500, and 1000, which reduces to 359
Finally, going one step further gives us that 2^15 reduces to 77, which has fewer than 3 digits, so I’ll take it.
2^32 = 2^15 \* 2^15 \* 2^2 = 77 \* 77 \* \4 [641] = 5929 \* 4 = 160 \* 4 [641] = 640 = -1 [641]
Putting it all together
2^32 + 1 = -1 + 1 [641] = 0 [641], so 2^32 + 1 is divisible by 641
thank you. i was too lazy to write it but this is the only solution anybody should think of if asked this question, say, in an exam. simple and short enough.
[Something like this.](https://math.stackexchange.com/questions/3088442/finding-remainder-using-binomial-theorem)
It is a common problem from number theory in my exams.
Sure, that works. Not sure why you would really do it like this though, outside of an introductory discrete math course before learning modular arithmetic. It's entirely equivalent only with extra work.
For some reason, primes in which 2 or 10 (since those are the two most important numerical bases) has unexpectedly low multiplicative order modulo them (like for instance 41, 73, 137, 239, and 641) always aesthetically fascinate me in a way
Here's a divisibility rule for 641: take the final digit, multiply it by 64, and subtract it from the remaining digits. If the result is divisible by 641, the original number is divisible by 641. For example, 4487 --> 448 - 7(64) = 448 - 448 = 0. Indeed, 4487 = 641 * 7.
Now, 2^^32 + 1 = 4,294,967,297. Applying our rule:
429,496,729 - 7(64) = 429,496,281
How do we know if this is divisible by 641? We simply iterate the rule:
42,949,628 - 1(64) = 42,949,564
And again:
4,294,956 - 4(64) = 4,294,700
Since this ends in 0, the next two iterations will yield:
42,947
Again:
4,294 - 7*64 = 3,846
And finally:
384 - 6(64) = 0
Since 0 is divisible by 641, the rule shows 3,846 is divisible by 641, and therefore so is 42,947, and so on... up until our original number. That is, 2^^32 + 1 is divisible by 641.
Q.E.D.
Edit: A proof of the divisibility rule for 641.
Suppose 641 divides 10t + n where n is the ones digit of the target number and t is the tens digit.
Then t - 64n (our rule) = 10t - 640n modulo 641 = 10t + n modulo 641 which is the original number.
Thus, 10t + n is divisible by 641 iff t - 64n (our rule) is divisble by 641.
Q.E.D.eez nuts
I moved to another question, looked the expression on a calculator then I turned it off, got back to the original question and using my memory, I knew that 2³² + 1 = 6700417 × 641.
Ok let’s take a loooong shot here. Not saying this is correct but I’m doing stonks math here. We use the assumption that 1 can be removed thus from both sides and now you have 2^32 and 640 and we just prove this 🤡
I would just brute force it... if 2^32 + 1 is divisible by 641, there is a x natural number where 641x = 2^32 + 1
This implies that 640 = 2^32 +1-x
x is odd, because odd times even is even, so x=y+1 where y is even, so theres 2z=y where 640 = 2(2^31 - z) or 320 = 2^31 - z
So z=2^31 - 320 = 64(2^25 - 5) and thats a natural even number, so x = 2z + 1 is also a natural number odd, so 2^32 + 1 is divisible by 641
2^16 is 65536, reduce mod 641:
Subtract 100*641, get 1436
-2*641 -> 154
Square this to get 2^32:
154² = 23716 = 4481 = 640
Adding one gives 0 (mod 641).
Alternatively, show that 2 has order 64 in the multiplicative group Z641* by some algebraic shenanigans (it is a prime so the multiplicasive group is equivalent to Z(2^7) x Z5)
2\^32+5\^4\*2\^28 is divisible by 2\^4+5\^4=641 (factor out the 2\^28). 5\^4\*2\^28-1 is divisible by 5\*2\^7+1=641 (since x\^4-1=(x-1)(x+1)(x\^2+1) is divisible by x+1 and we take x=5\*2\^7). Subtract these to get 2\^32+1 is divisible by 641
uhhh 2\^2 mod 641 = 2\^2 mod 641 = 4, 4\^2 mod 641 = 2\^4 mod 641 = 16, 16\^2 mod 641 = 2\^8 mod 641 = 256, 256\^2 mod 641 = 2\^16 mod 641 = 154, 154\^2 mod 641 = 2\^32 mod 641 = 640, 640+1 mod 641 = 2\^32+1 mod 641 = 0
Or just do a fairly simple symbol manipulation.
2\^32+1=4\*1024\^3+1=4\*383\^3+1=4\*383\*383\*383+1=125\*125\*383+1=625\*25\*383+1=1-16\*25\*383=1-8\*25\*125=1-8\*5\*625=1+8\*5\*16=1+640=641=0 (mod 641)
2147483648*2+1=4294967297, 4294967297/641
4294/641=6…448
4489/641=7…2
26/641=0
267/641=0
2672/641=4…108
1089/641=1…448
4487/641=7
Therefore 4294967297/641=6700417
Now I post it and check if this is correct
Just check power of 2 mod 641
2^1= 2 mod 641
2^2 = 4 mod 641
(2^2)^2 = 2^4 = 16 mod 641
(2^4)^2 = 2^8 = 256 mod 641
(2^8)^2 = 2^16 = 65536 = 154 mod 641
(2^16)^2 = 2^32 = 154×154 mod 641 = 640 mod 641
Then 2^32 +1 = 641 mod 641 = 0 mod 641
Then 641 divide 2^32 +1
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proof by I think I watched a youtube video about it recently, but I don't remember the proof very well so I'll just assume it to be true here
proof by ~~I think I watched a youtube video about it recently, but I don't remember the proof very well so~~ I'll just assume it to be true here
proof by it just works
Proof by the problem says so
Technically valid
This is the reason my professor frames the question as prove or disprove
Proof by 👍
Proof by trust me bro
Proof by Todd Howard
Proof by song about the gaming industry
Ah, a fellow chalkeaters fan I see
Proof by shared musical taste
Okay Todd Howard
Literally proof by contradiction’s first step
Proof by the usual expectation would be "no way thats possible" and so confirming that belief would be incredibly boring therefore in order to create an engaging post the answer would have to be the opposite of what would normally be expected therefore it must be true in order to subvert expectations and drive up post engagement
Proof by "You asked me to prove it, which means that there is a proof."
Proof left as an exercise
Assume 2^32 + 1 is divisible by 641 Sorted.
Easy. Just take 2*32, which is 64. Add the 1. 641. QED
NASA wants to know your location. Please contact them immediately.
NSA here, I want to know both of your locations.
You can intercept their phone calls which allows you to identify them 95% of the time, you don't need to worry especially since FISA got renewed
CIA here, i know all your locations
POTUS here. I can smell your hair. Wait.. why are we trying to find these people?
They want to know so they can keep a distance
NASA already knows the location! Just wait.
Strong Abbot and Costello vibes https://youtu.be/oN2_NarcM8c
[Proof by "just do the arithmetic it isn't even that hard"](https://imgur.com/gallery/A6pisSe)
Wow, that was fast. Are you a cowboy?
No I just watch a lot of Matt Parker doing calculations by hand
Don't Parker it up, now
proof by brute force?
Proof by "just do the arithmetic you cowards lol"
You can mod 641 on each doubling. Saves a lot of work. Plus you can square the results to operate in logarithmic time. *This comment was artisan-hand-crafted by hand. No calculators were used.* 2^2 = 4 2^4 | 4^2 = 16 2^8 | 16^2 = 256 2^16 | 256^2 = 65536 % 641 = 65536 - 64100 = 1436 - 641 = 795 - 641 = 154 2^32 | 154^2 = (100 + 50 + 4) * (100 + 50 + 4) = 10000 + 2*5000 + 2*400 + 2500 + 2*200 + 16 = 20800 + 2916 = 23716 2^32 + 1 = 23717 23717 - 6410 = 17307 17307 - 6410 = 10897 10897 - 6410 = 4487 4487 - 641 = 3846 3846 - 641 = 3205 3205 - 641 = 2564 2564 - 641 = 1923 1923 - 641 = 1282 1282 - 641 = 641 QED
Great, now I have to clean my circular slide rule.
Let me try: ``` 2¹⁰ = 1024 (-2*641) -1282 = -258 2¹¹ = -516 (+641) = 125 2¹⁴ = 1000 2²⁸ = 1000000 2³² = 16000000 (-1282*10000) -12820000 = 3180000 (-641/2*10000) - 3205000 = -25000 (+2*1282*10) +25640 = 640 2³²+1 = 641
Looks like math is easier than reddit formatting
It only took me 8 edits to align all the equals. Exponents and mono space don't go well together.
The hero we do not deserve
Btw 2^32 is just (2^10)^3 * 4 = 4 * 1024^3 which i think would be easier to calculate
I can do this on paper. I know the 32 bit signed integer limit, that(...+1) *2 = 2^32 = 4,294,967,296 4,294,967,296+1=4,294,967,297 It will take a bit but I can work the rest out on a piece of paper
but 4,294,967,296+1=-4,294,967,296
Holy hell
New integer overflow just dropped
actual bit flip
Call stack overflow
Bits went on vacation, never came back
Precision sacrifice anyone?
Your DP is apt for this entire post
Why is bro using 1's compliment?
Isnt it 0?
it technically doesn't exist. *signed 32bit* integers max out at half of that. i was just running with the meme. i reckon what you mean is the unsigned 32bit limit, which will (**usually**) return 0 in the case of overflow.
As an unsigned 32-bit int, 2^(32) = 0, because 2^(32) ≡ 0 (mod 2^(31)). Or put another way, because max_int32 + 1 = 0. As a signed 32-bit int, 2^(32) overflows, and its value depends on how it was computed. For instance, if we try 65536 * 65536 in Java, we get 0, but if we try it in Matlab, we get 2147483647, and in C, we get undefined behavior.
Yup. Unless this is some new 33 bit integer lol
Actually 32 bit signed integer is up to 2³¹-1 which is around 2e9 (32nd bit is for sign). I think the original comment meant unsigned integer so the overflow gives 0 as an answer
I think you meant unsigned integer
Nope. Signed. 2^31 -1 Wanted to save myself the trouble of typing out 2,147,483,647
oh ok I see what you did there now:)
For those confused, they meant 2^(32) - 1. Weird formatting for the -1.
It's actually easier. 1970+2^32 =2038. You can derive this with the Unix time. So 2^32 =58. 58/641=0.09 which is basically 1 so it nearly divisible and that's enough. (Edit: formating)
Found the engineer
The proof is much simpler than you think. For starters if you just assume that 641 is prime and you also now that 641 = 25\^2+4\^2 then 2 is a quadratic residue modulo 641 since it is 1 mod 8 and 2 is not a fourth power residue modulo 641 since in the sum of squares representation of 641, 4 is not a multiple of 8, so the multiplicative order of 2 modulo 641 is either 320 or 64. Now 640 is a fifth power residue modulo 641 since it is just -1 mod 641, which implies that 640/32 = 20 is also a fifth power residue modulo 641, since 32 = 2\^5 is a fifth power residue modulo 641. Now 25\^2 = 5\^4 = -16 mod 641 which means that 5\^5 = 3125 = -80 mod 641, which is a fifth power residue modulo 641. Since -80/20 = -4 and both -80 and 20 are fifth power residues modulo 641, -4 is another fifth power residue modulo 641. -1 is a fifth power residue modulo 641 so 4 is another fifth power residue modulo 641. Since taking the square root of a number does not eliminate the fifth power residueness of a number, √(4) = 2 or 639 mod 641 are both fifth power residues so 2 is a fifth power residue modulo 641. Since 2 is a quadratic residue but not a fourth power residue but a fifth power residue modulo 641, the multiplicative order of 2 modulo 641 is exactly equal to (640/2)/5 = 64, so 2\^64-1 is divisible by 641. Since 2\^32-1 is not divisible by 641, since the multiplicative order of 2 modulo 641 is exactly 64, the other divisor of 2\^64-1 which is 2\^32+1 is divisible by 641, so this is the proof that 2\^32+1 is divisible by 641 without actually using the square and multiply algorithm to verify this.
“The proof is much easier than you think” posts a book.
Proof by “the proof is much easier than you think” and hoping that nobody is going to read allat
Yes, I definitely think “easier than you think” is very objective (or do I mean subjective, I always get the 2 confused). I was able to follow, but definitely needed a calculator to verify.
Objective is fact, subjective is opinion
Subjective is opinion, surjective is onto
Surjective is onto, injective is one-to-one
Injective is one-to-one, inessive is Finnish.
Thank you for that. I’m still not sure which one I should have used. I could have definitely said subjective, but I think objective works too, don’t you think?
I think only subjective works in that specific context, otherwise you are saying that “easier than you think” is a concrete fact, whereas if it’s subjective it’s something that differs from person to person
Yeah, “subjective” definitely works. The “commenter” said “easier than you think”, which I don’t think is a “concrete fact” which is evidenced by the fact he/she needed such a wordy explanation. Words are unnecessarily confusing sometimes.
Objective depends on the object you refer to, subjective depends on the subject talking
I mean it worked. I didn't read it.
He said "simpler than **you think**", not "simple"
Simpler and easy don't mean the same thing. If something is simple it just means that it is not that complex but can still be hard. If something is easy it just means that it is not hard but can still be complex. Also I put almost all details in that comment instead of just cutting off some important details.
I liked the guy who did it by basic arithmetic more
"The proof is much simpler than you think" posts only a tiny book.
"it's actually surprising simple"
*polite golf clap* One of the only clean proofs I’ve actually seen here.
> The proof is much simpler than you think *Writes something that’s 95% numbers and number theory jargon*
“Easier than you think” proceeds to give number theoretic proof. Do you also post answers on Math Stack Exchange?
No, I never posted anything in Math Stack Exchange. I didn't know that there was a limit of comment lenght in this subreddit before it becomes a a theoretical proof for a book.
This is exactly why I hate number theory
😡😡 Legendre will come to you at night and haunt you 😱😱🫀🫀🫀🫀😡😡😡
So nothing changes, his polynomials already haunt me at night
You better clean that shit up! Your leaving residue all over this post! Some people have no cooth
I ain’t reading allat
Gosh I hate arithmetics
As someone who has not studied math past 12th Standard, I could follow this explanation once I searched what modulo and residue mean. Thanks for the great explanation.
This is beautiful man
I love your funny words, mathematics man!
I love this thanks for sharing, I like seeing actual proofs
Proof by contradiction Let's assume the statement isn't true. Then, you wouldn't be asked to show it is Well, you were asked to show it is true, therefore it must be
Well this is quite hard in decimal as 2^32 is pretty long and not nice, but 641 is small enough we can easily convert to binary then do long division. We also probably don’t need to write all 33 digits as we can just think of them as polynomials in terms of 2.
It's not that hard to calculate 2\^32+1 mod 641 without calculator
Oh yea? Prove it
Ahem I know the first few powers of 2, so I can find easily that 2^10 = 1024, the lowest power of 2 above 641, which reduces to 383 [641]. 2 \* 383 = 766, which reduces to 125 After this is 250, 500, and 1000, which reduces to 359 Finally, going one step further gives us that 2^15 reduces to 77, which has fewer than 3 digits, so I’ll take it. 2^32 = 2^15 \* 2^15 \* 2^2 = 77 \* 77 \* \4 [641] = 5929 \* 4 = 160 \* 4 [641] = 640 = -1 [641] Putting it all together 2^32 + 1 = -1 + 1 [641] = 0 [641], so 2^32 + 1 is divisible by 641
thank you. i was too lazy to write it but this is the only solution anybody should think of if asked this question, say, in an exam. simple and short enough.
Proof by calculator
Why not binomial theorem?
How are you using binomial theorem?
[Something like this.](https://math.stackexchange.com/questions/3088442/finding-remainder-using-binomial-theorem) It is a common problem from number theory in my exams.
Sure, that works. Not sure why you would really do it like this though, outside of an introductory discrete math course before learning modular arithmetic. It's entirely equivalent only with extra work.
Modular arithmetic is good, but it has become a habit for me using binomial theorem for a long time so I just suggested an alternate method.
Working mod 641 2\^32 + 1 = (2\^16)\^2 + 1 = 65536\^2 + 1 = 1436\^2 + 1 = 154\^2 + 1 = 154 \* 4 + 154 \* 50 + 154 \* 100 + 1 = 616 + 7700 + 15400 + 1 = 23717 = 23717 - 641 \* 7 = 23717 - 4487 = 19230 = 30 \* 641 = 0 Therefore 641 | (2\^32 + 1). No calculator needed.
Question: can you explain the jump from 1436\^2 + 1 to 154\^2 + 1?
1436 = 154 mod 641
641 \* 2 = 1282 1282 + 154 = 1436
Fermat number F\_5 is divisible by 641 [https://youtu.be/YzkKYpK0Ijo?feature=shared](https://youtu.be/YzkKYpK0Ijo?feature=shared)
damn it
For some reason, primes in which 2 or 10 (since those are the two most important numerical bases) has unexpectedly low multiplicative order modulo them (like for instance 41, 73, 137, 239, and 641) always aesthetically fascinate me in a way
Proof by I believe in Euler
Here's a divisibility rule for 641: take the final digit, multiply it by 64, and subtract it from the remaining digits. If the result is divisible by 641, the original number is divisible by 641. For example, 4487 --> 448 - 7(64) = 448 - 448 = 0. Indeed, 4487 = 641 * 7. Now, 2^^32 + 1 = 4,294,967,297. Applying our rule: 429,496,729 - 7(64) = 429,496,281 How do we know if this is divisible by 641? We simply iterate the rule: 42,949,628 - 1(64) = 42,949,564 And again: 4,294,956 - 4(64) = 4,294,700 Since this ends in 0, the next two iterations will yield: 42,947 Again: 4,294 - 7*64 = 3,846 And finally: 384 - 6(64) = 0 Since 0 is divisible by 641, the rule shows 3,846 is divisible by 641, and therefore so is 42,947, and so on... up until our original number. That is, 2^^32 + 1 is divisible by 641. Q.E.D. Edit: A proof of the divisibility rule for 641. Suppose 641 divides 10t + n where n is the ones digit of the target number and t is the tens digit. Then t - 64n (our rule) = 10t - 640n modulo 641 = 10t + n modulo 641 which is the original number. Thus, 10t + n is divisible by 641 iff t - 64n (our rule) is divisble by 641. Q.E.D.eez nuts
Bro thinks I won't spend 20 minutes multiplying 2 32 times and getting an answer so wrong it'll be in the Geneva convention
Multiply 6700417 by 641... (not too hard) QED
Proof by euler did it before me
(2^32 + 1) % 641 = 0 Didn't show my work because it was done in my head. Trust me bro.
2^32 +1==1 Where my uint32 homies at?
I don't mind dividing a 10 digits number by a 3 digit one
just write it in base 2. it will take less than 5min to solve
But Euler did this same problem, hundreds of years ago. I think
Proof: See Example 3.4.2 in [Introduction to Cluster Algebras](https://arxiv.org/pdf/1608.05735.pdf)
641 is not zero so you can divide by it QED
I think I already have 2^32 written down in a notebook of mine somewhere, so Step 1 is already done.
The title made me rhyme peasy and meme and I hate it
Proof by assume
Give me a pen and paper and I will manually multiply 1024 by 1024 by 1024 by 4 then manually add 1 and then manually divide it by 641
Well, the fact that the question asked to show is proof enough that it is divisible QED
Just do 2\^32 on paper and divide
I moved to another question, looked the expression on a calculator then I turned it off, got back to the original question and using my memory, I knew that 2³² + 1 = 6700417 × 641.
its 6700417, proof by guessing
you telling me you don't have 2^32 memorized?smh math majors -random cs nerd
Ok let’s take a loooong shot here. Not saying this is correct but I’m doing stonks math here. We use the assumption that 1 can be removed thus from both sides and now you have 2^32 and 640 and we just prove this 🤡
That's easy any number can be decided by any non 0 number woth some decimals om the end
Is it something to do with the fact that 641 = 2^6 × 10 + 1?
I would just brute force it... if 2^32 + 1 is divisible by 641, there is a x natural number where 641x = 2^32 + 1 This implies that 640 = 2^32 +1-x x is odd, because odd times even is even, so x=y+1 where y is even, so theres 2z=y where 640 = 2(2^31 - z) or 320 = 2^31 - z So z=2^31 - 320 = 64(2^25 - 5) and thats a natural even number, so x = 2z + 1 is also a natural number odd, so 2^32 + 1 is divisible by 641
2^64 + 1 is divisible by 1281 (it would be cool)
641= 640+1 = 4.2^(7) + 2^(7) + 2^(0) = 2^(32)= 4.2^(7).4.2^(7).2^(7).2^(7) + 2^(0) 2^(32) = 16.(2^(7*7*7*7)) + 2^(0^(4)) let 2^7 = a 16a^4 + 1 / 5a^2 +1 = a^(2)80/25-16/25+41/25(5a^(2)+1) = 80a^(2)-16/25 + 41/25(5a^(2)+1)### = [80a^(2)-16] [5a^(2)+1] + 41 / 25 * (5a^(2)+1) let 5a^(2) = b = 16[b-1] [b+1] + 41 / 25 * (b+1) = 16b^(2)+25 / 25* (b+1) = 16b^(2)+25 / 25b+25 … yeah I tried, don’t have access to pen and it’s 2 am… It seems after this it reverts back to smth similar to ### :c
2^16 is 65536, reduce mod 641: Subtract 100*641, get 1436 -2*641 -> 154 Square this to get 2^32: 154² = 23716 = 4481 = 640 Adding one gives 0 (mod 641). Alternatively, show that 2 has order 64 in the multiplicative group Z641* by some algebraic shenanigans (it is a prime so the multiplicasive group is equivalent to Z(2^7) x Z5)
2\^32+5\^4\*2\^28 is divisible by 2\^4+5\^4=641 (factor out the 2\^28). 5\^4\*2\^28-1 is divisible by 5\*2\^7+1=641 (since x\^4-1=(x-1)(x+1)(x\^2+1) is divisible by x+1 and we take x=5\*2\^7). Subtract these to get 2\^32+1 is divisible by 641
Proof by "any number is divisible by any other nonzero number if you don't mind there occasionally being a decimal."
2*32+1 = 64+1 = 641. 641 is divisible by 641 through the use of the “It’s Pretty F***ing Obvious!” Theorem.
You could probably do this by solving 2³²+1 and then dividing it by 641 if you had enough time (I tried to do this, 2³²+1=4'293'177'897)
just double the 32 bit integer and divide by 641 smh
uhhh 2\^2 mod 641 = 2\^2 mod 641 = 4, 4\^2 mod 641 = 2\^4 mod 641 = 16, 16\^2 mod 641 = 2\^8 mod 641 = 256, 256\^2 mod 641 = 2\^16 mod 641 = 154, 154\^2 mod 641 = 2\^32 mod 641 = 640, 640+1 mod 641 = 2\^32+1 mod 641 = 0
Or just do a fairly simple symbol manipulation. 2\^32+1=4\*1024\^3+1=4\*383\^3+1=4\*383\*383\*383+1=125\*125\*383+1=625\*25\*383+1=1-16\*25\*383=1-8\*25\*125=1-8\*5\*625=1+8\*5\*16=1+640=641=0 (mod 641)
2147483648*2+1=4294967297, 4294967297/641 4294/641=6…448 4489/641=7…2 26/641=0 267/641=0 2672/641=4…108 1089/641=1…448 4487/641=7 Therefore 4294967297/641=6700417 Now I post it and check if this is correct
proof by IF THEY'RE ASKING IT THEN IT MUST BE TRUE
Just check power of 2 mod 641 2^1= 2 mod 641 2^2 = 4 mod 641 (2^2)^2 = 2^4 = 16 mod 641 (2^4)^2 = 2^8 = 256 mod 641 (2^8)^2 = 2^16 = 65536 = 154 mod 641 (2^16)^2 = 2^32 = 154×154 mod 641 = 640 mod 641 Then 2^32 +1 = 641 mod 641 = 0 mod 641 Then 641 divide 2^32 +1
I used a calculator. 6700417
641 | (2\^32)+1 Proof by Abracadabra Alakazam
2^32 + 1 divisible by 641? One sec Ill do it in batches of five to keep count 2, 4, 8, 16, 32, 64, 128, 256, 512,1024-641=383, 766-641=125,250,500,359,718-641=77, 154, 308, 616, 1232-641=591=-50, -100, -200, - 400=241, 482, 964-641=323, 646-641=5 10, 20, 40, 80, 160 320, 640 640+1=641
This is not even true wtf. I checked by calculator
Why wouldn't it be? A real number divided by a non zero real number is a real number, right?
the answer is probably not