Fermat's Last theorem and its consequences

Years ago, French Mathematician Pierre Fermat stated his famous Last Theorem, that says, whereas the equation x^2 + y^2 = z^2 has an infinite no. of solutions, any equation of the type x^n + y^n = z^n where n > 2 has no solution whatsoever.

Many approaches were made, after which Euler was able to construct decent proofs for both n = 3 and n = 4. He was followed by Dirichlet, who constructed another decent proof for n = 5.

At this point, a general approach was made using a well-known property of numbers. It states that the equation a^2 - b^2 = c^2 has no integral solution of its own, which is easy to show from factorization. Thus for any even no. n, the theorem was firmly disproved because, for n=2m the equation can be re-written as (z^m)^2 - (x^m)^2 = (y^m)^2. Thus it only remained to disprove the theorem for odd primes. Further approaches were carried on, it continues even now-a-days.

I myself tried to disprove the theorem for n = 3, by using simple number theory approaches and found that it becomes extremely laborious and difficult. For readers' digest, I present a glimpse of my approach.

We are considered with the equation x^3 + y^3 = z^3. Via factorization, we rewrite the equation as,

x^3 + y^3 = z^3
=> (x + y)(x^2 + y^2 - xy) = z^3

so, the possible cases are:

1) (x + y) = kz
   (x^2 + y^2 - xy) = z^2/k

2) (x + y) = z^2/k
   (x^2 + y^2 - xy) = kz

3) (x + y) = kz^2
   (x^2 + y^2 - xy) = z/k

4) (x + y)=  z/k
   (x^2 + y^2 - xy) = kz^2

since (x^2 + y^2 - xy) > (x + y) for x,y > 2, so the 3rd case is impossible(since kz^2 < z/k so it claims k^2z < 1, a contradiction). We try to carry on with our remaining cases.

So,the first case follows like this,

x + y = kz
=> x^2 + y^2 + 2xy = k^2.z^2
=> 3xy = z^2(k^2 - 1/k) [subtracting the two equations from one another]
=> xy = z^2(k^3 - 1)/3k

To claim that the R.H.S. must be integer, we must have 3k | (k^3 - 1) or 3k | z^2.
But k does not divide (k^3 - 1) so the first possibility is discarded.

Thus, z^2 = 3^(2n).p^2.k^2 [where p contains no more factor of 3]
so, xy = 3^(2n - 1).p^2k.(k^3 - 1)

Finally, (x - y)^2 = (3^n.p.k^2)^2 - 4.3.3^(2n-2).p^2k[k^3-1]
=> (x - y)^2 = 3^(2n - 2).p^2[9k^4 - 12k(k^3 - 1)]

Thus, it reduces to the fact that, we must have k^4 - 12k(k^3 - 1) = some perfect square
or, k(12 - 3k^3) = some perfect square

which is possible only for k = 1, since for k > 1, 12 < 3.k^3. At k = 1 it is (12 - 3) = 9 = a perfect square.
Thus, x + y = z
x^2 + y^2 - xy = z^2 = (x + y)^2 = x^2 + y^2 + 2xy

which leads to xy = 0, a contradiction. Thus such a case is not obtainable in practical.
The other two cases will similarly lead to some sort of contradiction. But I think I already bored you enough of these approaches! So I stop it here...you can surf the net to get lots of modern techniques used to show the results for n=3,5...etc.

I haven't found the proofs constructed by Euler and Dirichlet. If someone knows any link on them, please comment below...

Monstrous Multiplications!!

Suppose you are taken in a room without any calculators, you are given 1 minute...and you are asked to calculate (257368 x 999999), what would you do??? Yeah, I am not joking...Vedic pundits were capable of doing such Monstrous multiplications within moments...right now I can tell the answer is 257367742632(since I know the technique), and once you know it too, you can tell answers of even bigger monsters, like (2418564 x 9999999) without sparing a minute!

The logic is simple, we are yet again about to utter the word "Complements"! If you observe, it is evident that 999999 = 1000000 - 1.

Thus, (257368 x 999999)
= 257368000000 - 257368
= 257367000000 + (100000-257368)
= 257367000000 + (complement of 257368)

since complement of 257368 comes out to be 742632, so, our result becomes 257367742632.

So, when a number is being multiplied by the biggest number of the same digit, to get the answer, first take the complement of the number, now subtract 1 from the number, and write the one-subtracted number and the complement side by side. The resulting number is the answer of the multiplication!

This is just an instance of the lightning-fast speed of calculation at that era. There are still a lot more to come! Vedic pundits referred to some more Sutra-s in this relation, for further ease of multiplications by uncommon numbers. I will discuss each of those Sutra-s with proper examples...

Subtraction in those early days

What do you do when someone asks you to subtract 723 from 1081?? Since 3>1, for the unit's digit subtraction, you take (1+10-3)=8, and turn the ten's digit from 8 to 7(unless u know some even easier technique of subtraction!). This two operation are known as Difference and Borrow. But at the ancient times, Aryan Pundits had no such concepts of difference and borrow, yet they used to subtract with acute perfection. Well, as I told earlier, all their computations were based on the two earlier definitions, and subtraction was not an exception either!

Let us recall the Sutra I stated a little ago, that is:

"निखिलं नवतश्चरमम दशतः"

Which means, "All from nine and last from ten". So, we are going to use our pet funda of complements.
I am presenting the process point wise,

1) We take our previous example. (1081-723). In this case, we refer the digits of 1081 to be Upper digits, and the digits of 723 to Lower digits, just for easier representation.

2) In each column of subtraction(by column I refer to unit's digit,ten's digit etc.), we take the difference of upper digit and lower digit, i.e. (upper digit ~ lower digit).

3) If in some column, upper digit > lower digit, then the answer becomes (upper digit - lower digit - 1). This extra 1 is subtracted since we are coming out of complements.

4) If in some column, upper digit < lower digit, the answer becomes the complement of (lower digit - upper digit).


Let's now take our example. I am referring unit's digit as column 1, ten's digit as column 2,...etc. So, in column 1, upper digit < lower digit, which means we are to take complement. Answer for column 1 is [10-(3-1)]=8.

For the next column, upper digit > lower digit, so we come out of complements, and the answer is (8-2-1)=5.

For 3rd column, upper digit < lower digit, so we need complement. Thus, the answer is [10-(7-0)]=3.

Finally for the last column, upper digit > lower digit, so the answer becomes, (1-0-1)=0.

So, our final answer is, 0358, i.e. 358. Clearly (358+723)=1081. So, our concept of complement proved to be useful once again.

Vedic Mathematicians intended to solve a lot of problems based on this complement concepts...one cannot imagine how powerful this technique can actually be for fast calculations!! But you will start believing in my words as soon as I illustrate the multiplication techniques used by them...

The two definitions

Computation at the Vedic era was entirely based on two easy definitions, one of them being Base, and other one being Complement.

Bases- The entire number system is built on the numbers 0-9. All of them keep repeating themselves in a specific order, but at some milestones. These "milestones" refer to 10,100,1000....etc. which we call ten's digit, hundred's digit....etc. in modern number theory. These numbers(which have "1" as their first digit, followed by zeroes) are termed as Bases.

Complements- Any number from the number line, when subtracted from its nearest base that is greater than it, gives the Complement of the number.

example: (i) Suppose we take 58, it is closest to the base 100, and 100 is greater than it. So its complement will be (100-52)=42.
 (ii) Suppose we have taken 42 instead of 58, in that case, 42 is closest to 10 rather than 100. But despite that we take the nearest base to be 100 since 10 is less than 42. Its complement is obviously 58.

# Obtaining Complement:

Like the Vedic Mathematicians defined Complements, they also suggested how to calculate complements of a number. This technique has been referred in a Sutra, which says...

"निखिलं नवतश्चरमम दशतः"

Which means, "All from nine and last from ten". If we observe carefully, we would notice that it only refers to the basic technique of subtracting higher digit numbers. Let us illustrate it with an example.

Suppose, we are to find the Complement of the number 1234. So, what we have to do is subtract the last digit from ten, i.e. (10-4)=6.
And the rest of the digits from 9, i.e. (9-1)=8, (9-2)=7, (9-3)=6. So the complement becomes 8766. Clearly (1234+8766)=10000.

A point to remember, when the number will end with a zero, i.e. 4810, then the technique is a bit different. We exclude the zero from the number first. i.e. the number becomes 481. Now, we determine its complement following the Sutra, which is found to be 519. Finally we put the zero at the end, and the number that formed is the complement of 4810, that is 5190.
If more than one zeroes are present at the end, then we exclude all of them, determine the complement, and finally put them back at the end.

The proof is even more simple, it is the "borrow" operation that reduces the base from 10 to 9 in each of the subtractions other than the last one. This concept of Complements was used in subtraction of two numbers, which I will discuss in the next article.