famous problems and famous mathematicians

You and your research or you and your studies for competitive math exams

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You and your research ( You and your studies) : By Richard Hamming, AT and T, Bell Labs mathematician;

In bubbles, she sees a mathematical universe: Abel Laureate, Prof Karen Uhlenbeck

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I was just skimming the biography “A Beautiful Mind” by Sylvia Nasar, about the life of mathematical genius, John Nash, Economics Nobel Laureate (and later Abel Laureate)…

Some math wisdom came to my mind: Good mathematicians look for analogies between theorems but the very best of them look for analogies within analogies; I was reading the following from the biography of John Nash: …It was the great HUngarian-born polymath John von Neumann who first recognized that social behaviour could be analyzed as games. Von Neumann’s 1928 article on parlour games was the first successful attempt to derive logical and mathematical rules about rivalries. Just as Blake saw the universe in a grain of sand, great scientists have often looked for clues to vast and complex problems in the small, familiar phenomena of daily life. Isaac Newton reached insights about the heavens by juggling wooden balls. Einstein contemplated a boat paddling upriver. Von Neumann pondered the game of poker.

\vdots

And, read how first woman Abel Laureate, genius mathematician Prof Karen Uhlenbeck sees a mathematical universe in bubbles…

Hats off to Prof Karen Uhlenbeck and NY Times author, Siobhan Roberts!!

Prof. Tim Gowers’ on recognising countable sets

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https://gowers.wordpress.com/2008/07/30/recognising-countable-sets/

Thanks Dr. Gowers’. These are invaluable insights into basics. Thanks for giving so much of your time.

Prof. Tim Gowers’ on functions, domains, etc.

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https://gowers.wordpress.com/2011/10/13/domains-codomains-ranges-images-preimages-inverse-images/

Thanks a lot Prof. Gowers! Math should be sans ambiguities as far as possible…!

I hope my students and readers can appreciate the details in this blog article of Prof. Gowers.

Regards,
Nalin Pithwa

Fermat-Kraitchik Factorization Method for factoring large numbers: training for RMO

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Reference: Elementary Number Theory, David M. Burton, 6th Edition.

In a fragment of a letter in all probability to Father Marin Mersenne in 1643, Fermat described a technique of his for factoring large numbers. This represented the first real improvement over the classical method of attempting to find a factor of n by dividing by all primes not exceeding \sqrt{n}. Fermat’s factorization scheme has at its heart the observation that the search for factors of an odd integer n (because powers of 2 are easily recognizable and may be removed at the outset, there is no loss in assuming that n is odd) is equivalent to obtaining integral solutions of x and y of the equation n = x^{2} - y^{2}.

If n is the difference of two squares, then it is apparent that n can be factored as n = x^{2}-y^{2} = (x+y)(x-y).

Conversely, when n has the factorization n=ab, with a \geq b \geq 1, then we may write n = (\frac{a+b}{2})^{2}-(\frac{a-b}{2})^{2}

Moreover, because n is taken to be an odd integer, a and b are themselves odd, hence, \frac{a+b}{2} and \frac{a-b}{2} will be nonnegative integers.

One begins the search for possible x and y satisfying the equation n=x^{2}-y^{2} or what is the same thing, the equation x^{2}-n=y^{2} by first determining the smallest integer k for which k^{2} \geq n. Now, look successively at the numbers k^{2}-n, (k+1)^{2}-n, (k+2)^{2}-n, (k+3)^{2}-n, \ldots until a value of m \geq n is found making m^{2}-n a square. The process cannot go on indefinitely, because we eventually arrive at (\frac{n+1}{2})^{2}-n=(\frac{n-1}{2})^{2} the representation of n corresponding to the trivial factorization n=n.1. If this point is reached without a square difference having been discovered earlier, then n has no other factors other than n and 1, in which case it is a prime.

Fermat used the procedure just described to factor 2027651281=44021.46061 in only 11 steps, as compared with making 4580 divisions by the odd primes up to 44021. This was probably a favourable case designed on purpose to show the chief virtue of this method: it does not require one to know all the primes less than \sqrt{n} to find factors of n.

\bf{Example}

To illustrate the application of Fermat’s method, let us factor the integer n=119143. From a table of squares, we find that 345^{2}<119143<346^{2}; thus it suffices to consider values of k^{2}-119143 for those k that satisfy the inequality 346 \leq k < (119143+1)/2=59572. The calculations begin as follows:

346^{2}-119143=119716-119143=573

347^{2}-119143=120409-119143=1266

348^{2}-119143=121104-119143=1961

349^{2}-119143=121801-119143=2658

350^{2}-119143=122500-119143=3357

351^{2}-119143=123201-119143=4058

352^{2}-119143=123904-119143=4761=69^{2}

This last line exhibits the factorization 119143=352^{2}-69^{2}=(352+69)(352-69)=421.283, where both the factors are prime. In only seven steps, we have obtained the prime factorization of the number 119143. Of course, one does not always fare so luckily — it may take many steps before a difference turns out to be a square.

Fermat’s method is most effective when the two factors of n are of nearly the same magnitude, for in this case, a suitable square will appear quickly. To illustrate, let us suppose that n=233449 is to be factored. The smallest square exceeding n is 154^{2} so that the sequence k^{2}-n starts with:

154^{2}-23449=23716-23449=267

155^{2}-23449=24025-23449=576=24^{2}. Hence, the factors of 23449 are 23449=(155+24)(155-24)=131

When examining the differences k^{2}-n as possible squares, many values can be immediately excluded by inspection of the final digits. We know, for instance, that a square must end in one of the six digits 0,1,4,5,6,9. This allows us to exclude all the values in the above example, save for 1266, 1961, 4761. By calculating the squares of the integers from 0 to 99 modulo 100, we see further that, for a square, the last two digits are limited to the following 22 possibilities:

00; 01, 04; 09; 16; 21; 24; 25; 29; 36; 41; 44; 49; 56; 61; 64; 69; 76; 81; 84; 89; 96.

The integer 1266 can be eliminated from consideration in this way. Because 61 is among the last two digits allowable in a square, it is only necessary to look at the numbers 1961 and 4761; the former is not a square, but 4761=69^{2}.

There is a generalization of Fermat’s factorization method that has been used with some success. Here, we look for distinct integers x and y such that x^{2}-y^{2} is a multiple of n rather than n itself, that is, x^{2} \equiv y^{2} \pmod {n}
.
Having obtained such integers d=gcd(x-y,n) (or, d=gcd(x+y,n)) can be calculated by means of the Euclidean Algorithm. Clearly, d is a divisor of n, but is it a non-trivial divisor? In other words, do we have 1<d<n?

In practice, n is usually the product of two primes p and q, with p<q so that d is equal to 1, p, q, or pq. Now, the congruence x^{2} \equiv y^{2} \pmod{n} translates into pq|(x-y)(x+y). Euclid's lemma tells us that p and q must divide one of the factors. If it happened that p|x-y and q|x-y, or expressed as a congruence x \equiv y \pmod{n}. Also, p|x+y and q|x+y yield x \equiv -y \pmod{n}. By seeking integers x and y satisfying x^{2} \equiv y^{2} \pmod{n}, where x \not\equiv \pm \pmod{n}, these two situations are ruled out. The result of all this is that d is either p or q, giving us a non-trivial divisor of n.

\bf{Example}

Suppose we wish to factor the positive integer n=2189 and happen to notice that 579^{2} \equiv 18^{2} \pmod{2189}. Then, we compute gcd(579-18,2189)=gcd(561,2189)=11 using the Euclidean Algorithm:

2189=3.561+506
561=1.506+55
506=9.55+11
55=5.11

This leads to the prime divisor 11 of 2189. The other factor, namely 199, can be obtained by observing that gcd(579+18,2189)=gcd(597,2189)=199

The reader might wonder how we ever arrived at a number, such as 579, whose square modulo 2189 also turns out to be a perfect square. In looking for squares close to multiples of 2189, it was observed that 81^{2} -3.2189 = -6 and 155^{2}-11.2189=-54 which translates into 81^{2} \equiv -2.3 \pmod{2189} and 155^{2} \equiv -2.3^{3} \pmod{2189}.

When these congruences are multiplied, they produce (81.155)^{2} \equiv (2.3^{2})^{2} \pmod{2189}. Because the product 81.155 = 12555 \equiv -579 \pmod{2189}, we ended up with the congruence 579^{2} \equiv 18^{2} \pmod{2189}.

The basis of our approach is to find several x_{i} having the property that each x_{i}^{2} is, modulo n, the product of small prime powers, and such that their product’s square is congruent to a perfect square.

When n has more than two prime factors, our factorization algorithm may still be applied; however, there is no guarantee that a particular solution of the congruence x^{2} \equiv y^{2} \pmod{n}, with x \not\equiv \pm \pmod{n} will result in a nontrivial divisor of n. Of course, the more solutions of this congruence that are available, the better the chance of finding the desired factors of n.

Our next example provides a considerably more efficient variant of this last factorization method. It was introduced by *Maurice Kraitchik* in the 1920’s and became the basis of such modern methods as the *quadratic sieve algorithm*.

\bf{Example}

Let n=12499 be the integer to be factored. The first square just larger than n is 112^{2} = 12544. So. we begin by considering the sequence of numbers x^{2}-n for x=112, 113, \ldots. As before, our interest is in obtaining a set of values x_{1}, x_{2}, x_{3}, \ldots x_{k} for which the product (x_{1}-n)(x_{2}-n)\ldots (x_{k}-n) is a square, say y^{2}. Then, (x_{1}x_{2}\ldots x_{k})^{2} \equiv y^{2} \pmod{n}, which might lead to a non-factor of n.

A short search reveals that 112^{2}-12499=45; 117^{2}-12499=1190; 121^{2}-12499=2142; or, written as congruences, 112^{2} \equiv 3^{2}.5 \pmod{12499} ; 117^{2} \equiv 2.5.7.17 \pmod{12499}; 121^{2} \equiv 2.3^{2}.7.17 \pmod{12499}. Multiplying these together results in the congruence: (112.117.121)^{2} \equiv (2.3^{2}.5.7.17)^{2} \pmod{12499}, that is, 1585584^{2} \equiv 10710^{2}\pmod{12499}. But, we are unlucky with this square combination. Because 1585584 \equiv 10710 \pmod{12499} only a trivial divisor of 12499 will be found. To be specific,

gcd(1585584+10710,21499)=1

gcd(1585584-10710,12499)=12499

After further calculation, we notice that

113^{2} \equiv 2.5.3^{3} \pmod{12499}

127^{2} \equiv 2.3.5.11^{2} \pmod{12499}

which gives rise to the congruence (113.127)^{2} \equiv (2.3^{2}.5.11)^{2} \pmod{12499}.

This reduce modulo 12499 to 1852^{2} \equiv 990^{2} \pmod{12499} and fortunately, 1852 \not\equiv \pm {990}\pm\pmod{12499}. Calculating

gcd(1852-990,12499)=gcd(862,12499)=431 produces the factorization 12499 =29.431

Problem to Practise:

Use Kraitchik’s method to factor the number 20437.

Cheers,
Nalin Pithwa

Wilson’s theorem and related problems in Elementary Number Theory for RMO

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I) Prove Wilson’s Theorem:

If p is a prime, then (p-1)! \equiv -1 {\pmod p}.

Proof:

The cases for primes 2 and 3 are clearly true.

Assume p>3

Suppose that a is any one of the p-1 positive integers 1,2,3, \ldots {p-1} and consider the linear congruence
ax \equiv 1 {\pmod p}. Then, gcd(a,p)=1.

Now, apply the following theorem: the linear congruence ax \equiv b {\pmod n} has a solution if and only if d|b, where d = gcd(a,b). If d|b, then it has d mutually incongruent solutions modulo n.

So, by the above theorem, the congruence here admits a unique solution modulo p; hence, there is a unique integer a^{'}, with 1 \leq a^{'} \leq p-1, satisfying aa^{'} \equiv 1 {\pmod p}.

Because p is prime, a = a^{'} if and only if a=1 or a=p-1. Indeed, the congruence a^{2} \equiv 1 {\pmod p} is equivalent to (a-1)(a+1) \equiv 0 {\pmod p}. Therefore, either a-1 \equiv 0 {\pmod p}, in which case a=1, or a+1 \equiv 0 {\pmod p}, in which case a=p-1.

If we omit the numbers 1 and p-1, the effect is to group the remaining integers 2,3, \ldots (p-2) into pairs a and a^{'}, where a \neq a^{'}, such that the product aa^{'} \equiv 1 {\pmod p}. When these (p-3)/2 congruences are multiplied together and the factors rearranged, we get

2.3. \ldots (p-2) \equiv 1 {\pmod p}

or rather

(p-2)! \equiv 1 {\pmod p}

Now multiply by p-1 to obtain the congruence

(p-1)! \equiv p-1 \equiv -1 {\pmod p}, which was desired to be proved.

An example to clarify the proof of Wilson’s theorem:

Specifically, let us take prime p=13. It is possible to divide the integers 2,3,4, \ldots, 11 into (p-3)/2=5 pairs, each product of which is congruent to 1 modulo 13. Let us write out these congruences explicitly as shown below:

2.7 \equiv 1 {\pmod {13}}
3.9 \equiv 1 {\pmod {13}}
4.10 \equiv 1 {\pmod {13}}
5.8 \equiv 1 {\pmod {13}}
6.11 \equiv 1 {\pmod {13}}

Multpilying these congruences gives the result 11! = (2.7)(3.9)(4.10)(5.8)(6.11) \equiv 1 {\pmod {13}}

and as 12! \equiv 12 \equiv -1 {\pmod {13}}

Thus, (p-1)! \equiv -1 {\pmod p} with prime p=13.

Further:

The converse to Wilson’s theorem is also true. If (n-1)! \equiv -1 {\pmod n}, then n must be prime. For, if n is not a prime, then n has a divisor d with 1 1 is prime if and only if (n-1)! \equiv -1 {\pmod n}. Unfortunately, this test is of more theoretical than practical interest because as n increases, (n-1)! rapidly becomes unmanageable in size.

Let us illustrate an application of Wilson’s theorem to the study of quadratic congruences{ What we mean by quadratic congruence is a congruence of the form ax^{2}+bx+c \equiv 0 {\pmod n}, with a \not\equiv 0 {\pmod n} }

Theorem: The quadratic congruence x^{2}+1 \equiv 0 {\pmod p}, where p is an odd prime, has a solution if and only if p \equiv 1 {\mod 4}.

Proof:

Let a be any solution of x^{2}+1 \equiv 0 {\pmod p} so that a^{2} \equiv -1 {\pmod p}. Because p \not |a, the outcome of applying Fermat’s Little Theorem is

1 \equiv a^{p-1} \equiv (a^{2})^{(p-1)/2} \equiv (-1)^{(p-1)/2} {\pmod p}

The possibility that p=4k+3 for some k does not arise. If it did, we would have

(-1)^{(p-1)/2} = (-1)^{2k+1} = -1

Hence, 1 \equiv -1 {\pmod p}. The net result of this is that p|2, which is clearly false. Therefore, p must be of the form 4k+1.

Now, for the opposite direction. In the product

(p-1)! = 1.2 \ldots \frac{p-1}{2} \frac{p+1}{2} \ldots (p-2)(p-1)

we have the congruences

p-1 \equiv -1 {\pmod p}
p-2 \equiv -2 {\pmod p}
p-3 \equiv -3 {\pmod p}
\vdots
\frac{p+1}{2} \equiv - \frac{p-1}{2} {\pmod p}

Rearranging the factors produces
(p-1)! \equiv 1.(-1).2.(-2) \ldots \frac{p-1}{2}. (-\frac{p-1}{2}) {\pmod p} \equiv (-1)^{(p-1)/2}(.2. \ldots \frac{p-1}{2})^{2}{\pmod p}

because there are (p-1)/2 minus signs involved. It is at this point that Wilson’s theorem can be brought to bear; for, (p-1)! \equiv -1 {\pmod p}, hence,

-1 \equiv (-1)^{(p-1)/2}((\frac{p-1}{2})!)^{2} {\pmod p}

If we assume that p is of the form 4k+1, then (-1)^{(p-1)/2} =1, leaving us with the congruence

-1 \equiv (-\frac{p-1}{2})^{2}{\pmod p}.

The conclusion is that the integer (\frac{p-1}{2})! satisfies the quadratic congruence x^{2}+1 \equiv 0 {\pmod p}.

Let us take a look at an actual example, say, the case p=13, which is a prime of the form 4k+1. Here, we have \frac{p-1}{2}=6, and it is easy to see that 6! = 720 \equiv 5 {\pmod {13}} and 5^{2}+1 = 26 \equiv 0 {\pmod {13}}.

Thus, the assertion that ((p-1)!)^{2}+1 \equiv 0 {\pmod p} is correct for p=13.

Wilson’s theorem implies that there exists an infinitude of composite numbers of the form n!+1. On the other hand, it is an open question whether n!+1 is prime for infinitely many values of n. Refer, for example:

https://math.stackexchange.com/questions/949520/are-there-infinitely-many-primes-of-the-form-n1

More later! Happy churnings of number theory!
Regards,
Nalin Pithwa