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Mohan’s generalized 50-cent theorem, a new step in number theory

September 17, 2011

Mohan, mdashf

Nature may speak mathematics but it’s often quiet …

Mohan’s generalized 50-cent theorem: f(m,k)=5*m+P=n^k, 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 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.

I will give a history of how i came to discover this an why I call it by the name of 50cent a celebrated HipHop musician. I am a hiphop musician but nobody knows me.

Long time ago on twitter, I was following 50cents and another celebrity from Hindi-wood.

I tweeted “I am having a hard time deciding who to unfollow 50 cents or AK”. It was easy 50cents had already blocked me for sending him my hip-hop tunes “I want hot water, like it’s not water..(what’s it git to be? Deuterium??)”

Anupam Kher 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 came up with a mathematical theorem which I will describe in it’s special case, from where I developed the general case above. Lets talk about the general case here and gradually go backwards.

(as I am copying my material from facebook where I originally posted it all, I need to gradually move to a paper-book or blackboard-white board format but nobody seems to give me an employment)

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 between 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 you 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 reduction in steps, as we will see later.

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 ..

SO for check of 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/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 tesed using this magical method..

let’s use to Brahma-gupta Fibonacci theorem !! (this is still needs to be checked again to understand the implications)

(n^k + l^k)(n’^k + l’^k) = 5.5.(m+m*)(m’+m*) + 10.(m+m*+m’+m’*)P, 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 between their individual scope of -(k-1) to k-1. If we have various powers then the scopes and integers will change accordingly ..

modifn: we are not using to BF theorem or identity as such but has a similar form and in case of exact theorem this can be applied (easier than the general case)

correction/note above:
1. in one of the m’s above I missed a prime (‘). figure it out (I will post my handwritten later)
2. we pick integers randomly between -(k-1) and k-1 for large powers k, because we do not know which integer will mke it divisible by 5. In a random case perhaps luck will help us save huge computation. (it will also take us away from the close number that is divisible by 5 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 … ie if one number is not a 7th power it is certainly not a 3rd power !!


Mikael Franzen 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 🙂

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 reg. BF part … right now I have discovered that Pythagorean theorem is actually a special case of our 50cent theorem for power 2 …

Another way of saying my generalized 50cent 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.

coroll: it seems pytha-gorean theorem for integers is a special case of 50cent theorem for power 2.

my recent language anaysis has paved the way for proving pytha-gorean is actually not (necessarily) a Greek name but an “apaba-ansha” of original sanskrut: padas-konas = base-angle


(I will append to the end my former posts reg. 50cent theorem, in it’s special cases which were the one I discovered/invented first)

a perfect square has a mathematical form: a^2 = 5m+P (where for power 2, P = -1,0,1) therefore a^2+b^2 = 5m+P + 5m’+P’ = 5(m+m’)+P+P’ = 5n+Q = c^2, 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 accomodated by P,P’ and Q … This is the proof of Pythagorous theorem.

      • (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. bada = ism/school of thought/line of thinking or step/proof/deduction etc. ansha = part or portion. “a result of wrong analysis, wrong form of language etc”. note that I have many times substantiated in my lang. 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 (found also in languages such as Japanese) ra and na convertions/alternations are similarly found in many languages. SO pytha-gorean <> paja-konan = paja-konas = pada-konas, etc. It could also be noted that gorean can be attributed to the apaba-ansha source gunan = multiplication/arithmetic … so base-arithmetic/base-angle rules etc…

note the meaning of Brahmagupta (whose theorem I’ve linked to 50cent theorem above) I had already given this analysis few weeks ago: Brahma is BrahMas or BruMas = (big/top/important) * (brain/head/mind). Gupta is used to mean secret or hidden. (so what is meaning of upagupta?) but Gupta may mean “proved” here. I don’t know. Brahmagupta actually may mean apart fom it’s meaning as a name, this: “proved by top-class mathematical minds” …

“brumaskupadita” may have been the actual apaba-ansha source of bahmagupta, note in that case it makes perfect sense: because in “sanskruta” >> “padana” is; to prove as in prati-padan/pratipadit. it’s actually “patit” instead of “padit” becaue da is a heavier tone of ta. again “patit” comes from paja/pata a step/place/base/proof. SO my final words are “bru-mas-ku-pata-ta” bru = generator of top-class/big, mas=mind, ku=particle of/from (as in Japanese ni, but used in present day Indian languages, Hindi: Un ke, jis ka, Odia: nka, nkara, ku etc) pata = (noun) step, ta = (verb ending) ta as in gruhi’ta = grant’ed,
[“bru-mas-ku-pata-ta” = verified/proved by top minds ]


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:
  • ~* 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;
“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.

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;

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 …”

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