“Nature may speak mathematics but it’s often quiet … “
Mohan’s generalized 50-cent theorem: a new theorem involving prime numbers, a new step in number theory.
m, k are integers and P is any integer between -(k-1) to (k-1)
Or in words: The integer powers of any integer is a multiple of a prime number within a integer scope of the power-integer.
I have explored so-far only the prime number 5 for which I have verified in detail that this is valid. It has many implications including towards Fermat’s theorem and Pythagorean theorem and many more corollaries with great importance towards computational number theory.
History; ( jump to latter parts, if you like, some unnecessary casual conversations here )
I will give a history of how I came to discover this and why I call it by the name of 50-cent, a celebrated Hip Hop musician. I am a hip hop musician but nobody knows me.
(pun intended, not credit, I am not going to be a serious musician in my life, prove me wrong, but serious intellectual involvement often takes us away from pursuing such enjoyable pastimes as being a musician, a person should perhaps not be too versatile, but I may be totally out of perspective here, just a thought)
Long time ago on twitter, I was following 50-cents and another celebrity from Hindi-wood. (Hindi-wood is Bollywood, pun intended)
I tweeted ” I am having a hard time deciding who to unfollow 50 cents or AK “.
It was easy; 50 cents had already blocked me (thank you @mdashf) for sending him my hip-hop tunes “I want hot water, like it’s not water… ( — what’s it got to be? Deuterium?? ) “
[celebrity people have a lot less patience for such, right? They better be, I was only being humorous, it seemed there were a few celebrities on twitter who had taken me kindly and lightly and I had exercised immense restrain not to be foolhardy to any-one, although one fine night I decided to unfollow every celebrity from my twitter, except Amitabh Bachchan, who I unfollowed recently, he is a tremendous actor, whose first movie I watched was when I was few years old, Dharam-Veer with Dharmendra, the two were my favorite actors for a long time.
Thankfully no-one perhaps took me personally except 50 cent. It seemed 50 cent was often responding to my tweets on his line, eg one of his “no-one sending you to jail for having pot” when I had said some tweets on pot. In all honesty 50 cent as much a fan of his music I am was great. He would come and flirt every nights with numerous ladies, as I would be following the tweets. That was when I would be spending a lot of time online, during times like this, 1 am, 2 am.
Talking about how humorous I was on twitter back then one example was celebrity Shashi Tharur, who said something about his marriage and I tweeted “at-least you should be happy you have a wife, I am awake alone here”. Talking about such mischief, I joked, after having some liquor ” What should India be more careful about, mosquitoes biting an innocent drunk or drinking problems as cultural erosion” or some such thing.
My drinking is no more a serious involvement, I have only in last few years taken very infrequent amount of liquor, eg once in 6 month or a year or maximum 3 months, I hear no applaud, I used to drink a lot, if such matters to your interest, partially because that was a habit since my long struggles in Particle Physics days. I think if I would like to inspire youngsters about anything its not alcohol, nor it is anything against it, it would be “virtues of silence, hard-work, personal and social ethics and an ability to think critically“. Rest you will know by yourself. ]
Anupam Kher, a celebrated actor of Indian movies, especially Hindi Movies where he did a marvelous and famous role-play mostly as a comedian, said on his line “those who want to be rude go ahead and unfollow me“, I unfollowed all celebrities from my twitter … except myself, as twitter does not allow you to follow or unfollow yourself. It was no insult towards any celebrity. I would regularly unfollow because I have to follow new people.
So thats the history, when 50 cent blocked my account, it kind of struck me that I should come up with something other than hip hop to impress him. And it turn out my past love maths came up. I was running some numbers in head starting with 50, and see that they follow patterns.
I came up with a mathematical theorem which I will describe in it’s special case, from where I developed the general case, as described in the Introduction above. Lets talk about the general case here and gradually go backwards.
( — as I am copying my content/material from face-book where I originally posted it all, I need to gradually move to a paper-book-keeping or blackboard-whiteboard format but nobody seems to give me an employment, pun intended. )
From the above general form therefore; any integer is a perfect k-power integer if it is divisible by 5, after or before addition of any integer in the integer-scope of power k, namely: -(k-1) to k-1.
This perfectibility test is achievable by 2(k-1)+1 steps therefore, because; we keep on adding one integer and divide by 5.
In cases, such as a power that in itself is a power this brings in additional constraints therefore, leads to reduction in steps, as we will see latter.
SO instead of 7 steps of testing if an integer is a 4th power or not, we need 6 steps, because; the first step will decide if it’s a perfect square or not … for checking a 9th power one does not need 17 steps but 15 steps because; the first 2 steps will decide if each result is a perfect cube or not.
NOTE; one can use a randomized approach to take integers from list or scope of integers in -(k-1) to k-1 and go on adding it and dividing by 5 till one gets an integer. If one does not, after all steps are exhausted, it is not a perfect k-power. So, an integer is k–power-free or not, can be tested using this magical method.
Let’s use our results to Brahmagupta Fibonacci theorem, hence forward BF theorem also !! ( — this still needs to be checked again, to understand the implications )
as usual n, l, n’, l’ are all natural numbers and k is an integer power and the m’s are resulting integers after division by 5 before or after addition of an integer, in their individual scope of -(k-1) to k-1. If we have various powers then the scopes and integers will change accordingly.
modification: we are not using our results to BF theorem or identity as such but our result has a similar form and in case of exact theorem this can be applied (easier than the general case)
correction note for above:
1. in one of the m‘s above I missed a prime (‘). (lets) figure it out ( — I am thinking of availability of handwritten proof !!)
2. we pick integers randomly between -(k-1) and k-1 for large powers k, because; we do not know which integer will make it divisible by 5. **
** In a random case perhaps luck will help us save “huge” computation. It might also take us away from the closer number that is divisible by 5 if we are to chose randomly so we can also stick to a pattern from negative to positive, let’s say.
NOTE; one can use simultaneously different powers to check if the number is the next power or not … eg one may continue from 2nd power to 3rd power without losing steps … i.e. if one number is not a 7th power it is certainly not a 3rd power !!
Mikael Franzen ( — remarks ) I like it! Brahmagupta’s identity is a nice touch, but, I don’t quite follow on your use of the prime and asterisk symbols here. You mean set component, compliment … or something else? Explain please 🙂
My answer; m and m* and m’ are just different integers, that is all. Because we have n, l, etc. Also I need to check again for correctness regarding BF part … right now I discovered that Pythagorean Theorem is actually a special case of “our” 50 cent theorem for power 2 … [ Franzen and I were planning to publish this work together, we published one paper on Relativity together and a couple others were expected to be. The paper linked here was the first paper of 2012 on the U Texas archive.]
Our published paper on Relativity; Manmohan Dash, Mikael Franzen Time dilation have opposite signs in hemispheres of recession and approach (501 K, pdf) Jan 7, 12 Abstract , Paper (src), View paper (auto. generated pdf), Index of related papers.
Another way of saying the generalized 50 cent theorem: all perfect k–power of an integer , 9 is a perfect 2 power, 27 a perfect 3 power, k being an integer itself, with scope -(k-1) to k-1, are integral multiples of 5 before or after adding an integer from this scope.
corollary: it seems pythagorean theorem for integers is a special case of 50 cent theorem for power 2.
Linguistics of Pythagoras, Brahmagupta
Update ( — 16-12-2013 ) ; In the preliminary days of my linguistic analysis I used to take interesting words and see if phonetically the have any connection to what sense can be made in present days. Also after I continued doing such mindful analysis of interesting words gradually I discovered the most interesting idea of mine so far in my own analysis is, what I call Sun theory of language, or occasionally; X or M or XM or XaMa theory, for related reasons. Once that unification idea of Sun theory came up I haven’t analyzed back all the other interesting words such as this, written intermittently into various articles.
But its a given and can be readily seen in the context of any word in any language, that such words will merely in many different ways connect, phonetically, sometimes with little variations or elemental alternations, to: Sun, Sun’s attributes etc. (eg here Pythagora will go to satya–goura, shudha–goura etc) See below what I thought in the preliminary days of the linguistics analysis, before the unification and contingent nature of language was as clear to me as is today. Nonetheless the connections are valid, if not one-way exclusively.
My recent language analysis has paved the way for proving pytha-gorean is actually not ( — necessarily ) a Greek name but ( also possibly, and by all means) an “apaba–ansha” of original Sanskrit: padas–konas = base-angle
NOTE; ( — I will append to the end my former posts reg. 50 cent theorem, in it’s special cases which were the one I discovered first chronologically)
A perfect square has a mathematical form: a(square) = 5 m + P ( — where for power 2, P = -1,0,1) therefore a(square) + b(square) = 5 m + P + 5 m’ + P’ = 5 (m+m’) + P + P’ = 5 n + Q = c(square), n and Q are again integers and Q has a different scope than P. But a, b, c are not necessarily rational which may be accommodated by P, P’ and Q … This is the proof of Pythagoras Theorem. Note that a(square) is simply square of a.
- The language analysis I append here and we will proceed to the number theory after this.
NOTE; apaba–ansha = apabada–ansha = apabru–ansha etc. apa = wrong ( — a more refined analysis says apa = Indic: alpa = low/less, hence “less correct”, its same as Greek: ap as in apoapis) Similarly, bada (baadaw) = ism or school-of-thought or line-of-thinking or step, proof, deduction etc. ansha = part or portion. “a result of wrong analysis, wrong form of language or lesser correct etc“.
note that I have many times substantiated in my language analysis articles and applying here; pytha could be a result of pada, paja, pata = base or foot and ga is a heavier tone of ka ( — ga, ka found also in prominently phonetic languages such as Japanese ) ra and na conversions or alternations are similarly found in many languages. SO pytha-gorean <> paja-konan = paja-konas = pada-konas, etc. Where <> denotes alternations or phonetic conversions, it could also be noted that gorean can be attributed to the apaba-ansha source gunan = multiplication or arithmetic … so base-arithmetic or base-angle rules etc.
The phonetic alternations have preserved something like this actually: Arithmetic in Odia is; Pati-Ganita, algebra is; Bija Ganita. If you can see thats a directly phonetic matching with Pythagora. (needs r<>n, but also pati is Odia for mouth and arithmatics is supposed to be easier to keep by heart and bija means seeds, perhaps the x-y variables are seeds?)
note the meaning of Brahmagupta ( — of BF Theorem I’ve linked to 50 cent theorem above ) I had already given this analysis few weeks ago: Brahma is Brah+Mas or BruMas = (big, top, important) * (brain, head, mind). Gupta is used to mean secret or hidden. ( — so what is meaning of upagupta? see I asked that in September 2011 and in September 2013 I already had an answer from SUn Theory, Gupta i secret because its Sun, Sun is hidden half-the-day, upagupta merely says the Sun at the top. or the region at top, etc, upa comes from and goes into upper, hence uparisthagupta = upagupta. )
Gupta may mean “proved” here. I don’t know. Brahmagupta actually may mean apart from it’s meaning as a name, this: “proved by top-class mathematical minds, or proved by top methods” … Gupta altered by a p<>e and inheriting a r goes to gruhita which is Indic: accepted, proved, granted.
“brumaskupadita” may have been the actual apaba-ansha source of brahmagupta, note in that case it makes perfect sense: because in “sanskruta= Sanskrit” >> “padana” is; to prove as in prati-padan or pratipadit. it’s actually “patit” instead of “padit” becaue da is a heavier tone of ta. again “patit” comes from paja or pata a step, descend, fall, place, base, proof etc. SO my final words are “bru-mas-ku-pata-ta” bru = generator of top-class, big, mas = mind, ku = particle of or from (as in Japanese ni, but used in present day Indian languages, Hindi: Un ke, jis ka, Odia: n’ka, n’ka’ra, ku etc) pata = (noun) step, ta = (verb ending) ta as in gruhi’ta = grant’ed. (I gave above Gupta = gruhita without realizing that 2 years ago I had already given this example of gruhita = granted, amazing isn’t it? )
[“bru-mas-ku-pata-ta” = verified, proved by top minds/methods/steps ] One last note of language; in Indianic and Japanese or any phonetic language what you should be careful of, if you are using Roman script? THIS.
Now back to the history of the theorem, which I discovered in it’s special form, about a year ago, and the discussion of teh special case will follow after this.
50cents was flirting with women all the time on twitter, he seemed to be a humorous fella, tweeting to one of my tweets “they will not put you in jail for smoking a pot..”
50 cents irritated me so much, not because of this comment or flirting but he blocked my account when I wrote him a tweet with a hip-tune, I came up with the mathematical theorem, all perfect squares, + or – 1 are divisible by 5 …
50 – 1 = 49, = 7 squared .. right? Now check all the small integers plus random numbers, for power of two.
1^2 -1, 2^2 +1, 3^2 +1, 4^2 -1, 5^2, 6^2 -1, 7^2 +1, 8^2+1, 9^2-1, 10^2, 11^2 -1, 12^2+1, 13 ^2 +1, ……. 221^2-1, …. 55328^2 +1…
MY generalization*~*: you will always find intergers m and n such that a function f(m) = 5*m = n^2 + P, where P(0) =0, P(1) =1, P(2) = -1
I call the above “50cent theorem” for power two.
Mike (Franzen) has a generalization which I haven’t checked:
so for all such cases: ∃ x: P(x) = ±1 hence: x^n… = n^2 = ±1 (n − 1)2 + (n − 1) + n
where x^n= k=±1(sub)Σ, n(sup)Σ = (nk ±1 )
- Mohan’s hipster/50 cent theorem: the 5th multiple of any interger is a square of another integer differing at most by a step of 1.
- this means for very large numbers you don’t need to find it’s squre but the 5th multiple of another integer …
(In Japanese in RomanJi) >> dasu ga suji no atarashio (?) hosoku wa shomeideshita, sore wa kare wa “50 centa hosoku” namae de hanashita, kare wa kono kotae de totemo “tanoshii ureshii” omoimasu …
another way: of saying the theorem: all perfect squares (of integes) are multiples of 5. If not add or subtract a 1 after the square.
yet another way: the squares of all natural numbers are multiples of (divisible by) 5. In the cases it is not it differs only by 1.
“I do not know how many new theorems it takes to publish in maths !!” but we will link it to square free numbers (*~*) [Mike want me to publish this and be a coauthor: very good idea, but I do maths like poems who will read them?, SO I tell him: If you prove the above functional form of the equation by induction you are my co-author]
- corollary: if (5*m + P) / n^2 = 1, 5*m+P is not a squarefree number. [refer for definition of squarefree numbers: http://en.wikipedia.org/wiki/Square-free_integer%5D
- one more corollary: on wikipedia above, there is a code/boolean that decides if a number is a perfect square or not, now using my newly invented theorem you can do this: take any number, add, 0 or +/- 1 and in each 3 cases divide the number by 5. If divisible it’s a perfect square that is not a square-free number.
the code was;
if (n & 7 == 1) or (n & 31 == 4) or (n & 127 == 16) or (n & 191 == 0) then
return n is probably square
return n is definitely not square
My algorithm/code is;
if (n)/5 or (n+1)/5 or (n-1) /5 = k then
return ” nis probably square”
else “n is definitely not square”.
(the computer must know k is an inetger and n is an integer)
My note to Mike; “in this last hour you were away I have invented one of the most powerful mathematical theorems that I know. Check my last post, now it’s generalized to any integer power not just square and I have now defined a powerfree number as compared to a squre free number by checking to see if the number is divided by 5. This replaces complicated algorithms that check the squarity of numbers. Now we can check the “powerity” of any natural numbr by dividing it by 5, before or after addition of any integer between the scope of preceding integer of the power. (or by defn scope of the integer of power) I already checked by pen and paper and see that this easily follows if I recognized my P as an integer between the scope of preceding integer of the power. Also now computation *has* been made much easier using this theorem”
I add this refernce which I checked to believe actually my theorems and corrolaries haven’t beed discovered ever; http://2000clicks.com/mathhelp/NumberPerfectSquares.aspx
“Theorems and Factoids Involving Perfect Squares”
I CONFIRM: new mathematical theorem invented by me, this can be used to check the squrity of numbers in at most 3 steps of division and atmost 2 steps of addition/subtraction.
I think “it replaces previous complicated and computation intensive algorithm and mathemtical theorem” … if any, I gave example of the above
I will check for cubic numbers but need to prove this theorem by induction for squares …if you do this you are an author in my publication.
FOR POWERS OF TWO;
step1: take any number, *natural number or integer, howsoever big*, divide by 5, if returns n integer it is a possible perfect square
step2: if you did not get integer in last step add 1 and divide by 5, if returns integer it’s a possible perfect square
step3: if you did not get an integer in last step subtract 1 and divide by 5, if you get an integer it’s a possible perfect square.
If not it is not a perfect square, that is in these 3 steps we proved if a number is squarefree.
“found the formula for cubic numbers… yey”
cubic numbers and square numbers follow same formula… that I alreadt gave
NOTE; the generalization to 3 and all powers I established on September 1, 5 AM or earlier. The special case for power two I discovered about a year ago, exact date on my Twitter. I will see if I can recover this.
so i guess 4th exponent will also follow same pattern, the cubes will be confirmed in 5 steps, as compared to 3 steps of squares, the 4th exponent will perhaps be confirmed in 7 steps …
by the way the 4th exponent will be confirmed in at most 7 steps but before that one can check it must be a perfect square or not which will reduce the number of steps …
so perfect 4th powers are testable only in 6 steps
COROLLARY: I just hinted above that my 50cent theorem is a general case of a pythagorean (paja-konas. pada-konas) theorem. here is a complicated way to prove that the area of a right-angle triangle with integer base-height is not a perfect square. my simple proof using 50cent theorem will follow after this link
Pythagorean Square Triangle — a proof by infinite descent that the area of an integer-sided right triangle can’t be a square number
my proof: see how elegant 50cent theprem can be; 1/2 ab = 1/2 sqrt[(5m+P)(5m’+P’)] = 1/2 sqrt [ 5(5mm’ + mP’ + m’P) + PP’] = 1/2 sqrt [5m*+Q], clear that m* is integer and Q is integer with same scope as P and P’ hence correspond to power 2, so 5m*+Q is a perfect square. that means; 1/2 ab = 1/2 c. if c is a perfect square 1/2 is still not one. SO 1/2 ab can not be a perfect square …
NOTE; what mathematicians call Fermat’s last theorem or proof by infinite descent is a lengthy and cumbersome proof, basically I have thrown Fermat’s last theorem to prove area of right angle triangle to ocean by the above proof … here is Fermat’s lengthy proof if you still like it; http://fermatslasttheorem.blogspot.com/2005/05/fermats-one-proof.html
Fermat’s Last Theorem: Fermat’s One Proof
The purpose of this blog is to present the story behind Fermat’s Last Theorem and Wiles’ proof in a way accessible to the mathematical amateur.
what an invention I made last night (from my original discovery a year ago) now my 50cent theorem proves pythagorus and Fermat’s theorem.
and it proves many corollaries such as area of a right angle with integer base-height is not a perfect square and another corollary that gives a computer algorithm of finding for sure if any natural number in the Universe is a k-power-free or not where it is k-power free by definition if the number is not dvisible by a perfect k-power, k being a natural number integer itself … it’s a new mathematical paradigm I have invented last night …
Now you can throw Fermat’ proof by infinite descent to whereever you want. I do not know what proof was made in 1995, need to look for it …
the proof of Fermat’s theorem was given actually by Euler by “infinite descent” method. Fermat was a lazy guy (like me?) he did not like giving proofs (lucky that he has come true in so many cases). Anyway let’s see what my 50cent theorem is capable of doing …
my generalizations of 50cent theorem is not letting me sleep. I am quite doubtful of how necessarily and sufficiently it fits into number theory. eg the scope of integers defined from the integer-power (of an integer itself); what if a numbr lies outside this scope but still gives a “solution”.
Since I have proved a few corollaries of PG theorem, it’s still a cmputational tool because it can give the “probability” that a given number to be tested is powerfree or not …
the other theorems such as prime-factorization and Euclid’s algorithm (highest common factor ) can prove very useful. Also the functional form I have got for 50cent theorem has an integer which is a prime-factor multiple (5 is a prime number) the scope is always odd for even powers and even for odd powers etc …
in simplicity my 50cent theorem says this: you take any number (integer) that you want to check as a k-power of another integer. you add any integer between -(k-1) .., 0, .., (k-1) to the “test integer” one by one, each time you divide by 5, if the result is an integer it is probably a “perfect” power else definitely not. Once all the integers to be added have been added and no more is there you can claim it’s not a perfect power. SO this in-fact is a computational proof of Fermat’s theorem for any power.
this can be used as a code which tests any integers, if ever Fermat’s theorem is invalid for any integer (that is any integer set is found that is a solution to Fermat’s equation for higher powers, it can raise a flag) This code can be fitted to all the computers that compute integer powers …
so (9898)^20 + (667)^20 is evaluated and that number is added to -20, -19, -18, …0, …, 20 one by one. it’s a 41 steps, if in 41 steps of dividing by 5 we never reach an integer dividend we know that Fermat was probably right hence no integer raised to the power 20 equals to (9898)^20 + (667)^20 …
one can test 1000 equations with different powers, 8, 9, 20, whatever is possible with the computer … and get a null result …
note that this must also give positive result for Pythagorean theorem that is for Fermat’s theorem for power 2 of any two integers the sum will always give an integer upon dividing by 5 before or after adding/subtractiong: 0, 1
let me test a Pythagorean sample from mind: (well I already tested this for many numbers so let me chose a large number)
660^2 = 435600, 23^2 = 529. you can see that just adding 1 to 529 will give us a number which is divisible by 5. and the actual solution is 660.4006 which is not an integer, but that’s because we have a scope which is non-zero (-1, 0, 1) hence adds a fraction to any one side or all of them. SO the possibility that we have a perfect square is null despite division by 5 returning an integer. you can take 5 and 12 and it gives 13 as a solution, but this time 4^2 – 1 and 12^2 + 1 gives us numbers that are divisible by 5. But so does 13^2 … we are just lucky to find integral solution. But despite of a non-zero probability of finding a solution to Fermat’s higher power equations we never get a solution.
let’s check for 3rd power: 6^3 + 8^3 = 216 + 512 = 728, adding a 2 divides by 5 but we never get an integer cube root of 728 … (web-search says: 8.99588289055 )
WE need additional constraints and theorems to prove that Fermat theorem for higher power is true …
but there is a catch for my 50cent theorem for higher powers: we may always get away with divisibility of 5. May be we formulate interms of a higher prime factor, then perhaps with lesser scope we can see a number was in-fact a power-free number …
by defining a higher prime factor such as 13 we get a smaller and smller scope to be added to the test integer and in very few steps we can see it is not a perfect power, so my head ache is now gone …
the hint is make power tables and find a pattern-formula with a higher primate, take a smaller scope and reject non-perfectpowers from the data..
“nature may speak mathematics but it’s often quiet …”