# Chennai Mathematical Institute Entrance

# Trigonometric Telescopic Sums and Products: A free video lecture from Mathematics Hothouse

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

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 . 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 .

If n is the difference of two squares, then it is apparent that n can be factored as .

Conversely, when n has the factorization , with , then we may write

Moreover, because n is taken to be an odd integer, a and b are themselves odd, hence, and will be nonnegative integers.

One begins the search for possible x and y satisfying the equation or what is the same thing, the equation by first determining the smallest integer k for which . Now, look successively at the numbers , , , , until a value of is found making a square. The process cannot go on indefinitely, because we eventually arrive at the representation of n corresponding to the trivial factorization . 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 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 to find factors of n.

To illustrate the application of Fermat’s method, let us factor the integer . From a table of squares, we find that ; thus it suffices to consider values of for those k that satisfy the inequality . The calculations begin as follows:

This last line exhibits the factorization , 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 is to be factored. The smallest square exceeding n is so that the sequence starts with:

. Hence, the factors of 23449 are

When examining the differences 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 .

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 is a multiple of n rather than n itself, that is,

.

Having obtained such integers (or, ) 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 ?

In practice, n is usually the product of two primes p and q, with so that d is equal to 1, p, q, or pq. Now, the congruence translates into . Euclid's lemma tells us that p and q must divide one of the factors. If it happened that and , or expressed as a congruence . Also, and yield . By seeking integers x and y satisfying , where , 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.

Suppose we wish to factor the positive integer and happen to notice that . Then, we compute using the Euclidean Algorithm:

This leads to the prime divisor 11 of 2189. The other factor, namely 199, can be obtained by observing that

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 and which translates into and .

When these congruences are multiplied, they produce . Because the product , we ended up with the congruence .

The basis of our approach is to find several having the property that each 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 , with 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*.

Let be the integer to be factored. The first square just larger than n is . So. we begin by considering the sequence of numbers for . As before, our interest is in obtaining a set of values for which the product is a square, say . Then, , which might lead to a non-factor of n.

A short search reveals that ; ; ; or, written as congruences, ; ; . Multiplying these together results in the congruence: , that is, . But, we are unlucky with this square combination. Because only a trivial divisor of 12499 will be found. To be specific,

After further calculation, we notice that

which gives rise to the congruence .

This reduce modulo 12499 to and fortunately, . Calculating

produces the factorization

Problem to Practise:

Use Kraitchik’s method to factor the number 20437.

Cheers,

Nalin Pithwa

# Questions based on Wilson’s theorem for training for RMO

1(a) Find the remainder when is divided by 17.

1(b) Find the remainder when is divided by 29.

2: Determine whether 17 is a prime by deciding if

3: Arrange the integers 2,3,4, …, 21 in pairs a and b that satisfy .

4: Show that .

5a: Prove that an integer is prime if and only if .

5b: If n is a composite integer, show that , except when .

6: Given a prime number p, establish the congruence

7: If p is prime, prove that for any integer a, and

8: Find two odd primes for which the congruence holds.

9: Using Wilson’s theorem, prove that for any odd prime p:

10a: For a prime p of the form , prove that either

or

10b: Use the part (a) to show that if is prime, then the product of all the even integers less than p is congruent modulo p to either 1 or -1.

More later,

Nalin Pithwa.

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

I) Prove Wilson’s Theorem:

If p is a prime, then .

Proof:

The cases for primes 2 and 3 are clearly true.

Assume

Suppose that a is any one of the p-1 positive integers and consider the linear congruence

. Then, .

Now, apply the following theorem: the linear congruence has a solution if and only if , where . If , 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 , with , satisfying .

Because p is prime, if and only if or . Indeed, the congruence is equivalent to . Therefore, either , in which case , or , in which case .

If we omit the numbers 1 and p-1, the effect is to group the remaining integers into pairs and , where , such that the product . When these congruences are multiplied together and the factors rearranged, we get

or rather

Now multiply by p-1 to obtain the congruence

, which was desired to be proved.

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

Specifically, let us take prime . It is possible to divide the integers into pairs, each product of which is congruent to 1 modulo 13. Let us write out these congruences explicitly as shown below:

Multpilying these congruences gives the result

and as

Thus, with prime .

Further:

The converse to Wilson’s theorem is also true. If , then n must be prime. For, if n is not a prime, then n has a divisor d with is prime if and only if . Unfortunately, this test is of more theoretical than practical interest because as n increases, 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 , with }

Theorem: The quadratic congruence , where p is an odd prime, has a solution if and only if .

Proof:

Let a be any solution of so that . Because , the outcome of applying Fermat’s Little Theorem is

The possibility that for some k does not arise. If it did, we would have

Hence, . The net result of this is that , which is clearly false. Therefore, p must be of the form .

Now, for the opposite direction. In the product

we have the congruences

Rearranging the factors produces

because there are minus signs involved. It is at this point that Wilson’s theorem can be brought to bear; for, , hence,

If we assume that p is of the form , then , leaving us with the congruence

.

The conclusion is that the integer satisfies the quadratic congruence .

Let us take a look at an actual example, say, the case , which is a prime of the form . Here, we have , and it is easy to see that and .

Thus, the assertion that is correct for .

Wilson’s theorem implies that there exists an infinitude of composite numbers of the form . On the other hand, it is an open question whether 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

# Eight digit bank identification number and other problems of elementary number theory

Question 1:

Consider the eight-digit bank identification number , which is followed by a ninth check digit chosen to satisfy the congruence

(a) Obtain the check digits that should be appended to the two numbers 55382006 and 81372439.

(b) The bank identification number has an illegitimate fourth digit. Determine the value of the obscured digit.

Question 2:

(a) Find an integer having the remainders 1,2,5,5 when divided by 2, 3, 6, 12 respectively (Yih-hing, died 717)

(b) Find an integer having the remainders 2,3,4,5 when divided by 3,4,5,6 respectively (Bhaskara, born 1114)

(c) Find an integer having remainders 3,11,15 when divided by 10, 13, 17, respectively (Regiomontanus, 1436-1476)

Question 3:

Question 3:

Let denote the nth triangular number. For which values of n does divide

Hint: Because , it suffices to determine those n satisfying

Question 4:

Find the solutions of the system of congruences:

Question 5:

Obtain the two incongruent solutions modulo 210 of the system

Question 6:

Use Fermat’s Little Theorem to verify that 17 divides

Question 7:

(a) If , show that . Hint: From Fermat’s Little Theorem, and

(b) If , show that divides

(c) If , show that

Question 8:

Show that and . Do there exist infinitely many composite numbers n with the property that and ?

Question 9:

Prove that any integer of the form is an absolute pseudoprime if all three factors are prime; hence, is an absolute pseudoprime.

Question 10:

Prove that the quadratic congruence , where p is an odd prime, has a solution if and only if .

Note: By quadratic congruence is meant a congruence of the form with . This is the content of the above proof.

More later,

Nalin Pithwa.

# Pre RMO algebra : some tough problems

Question 1:

Find the cube root of

Question 2:

Find the square root of

Question 3:

Simplify (a):

Simplify (b):

Question 4:

Solve :

Question 5:

Solve the following simultaneous equations:

and

Question 6:

Simplify (a):

Simplify (b):

Question 7:

Find the HCF and LCM of the following algebraic expressions:

and and

Question 8:

Simplify the following using two different approaches:

Question 9:

Solve the following simultaneous equations:

Slatex x^{2}y^{2} + 192 = 28xy$ and

Question 10:

If a, b, c are in HP, then show that

Question 11:

if , prove that

Question 12:

Determine the ratio if we know that

More later,

Nalin Pithwa

Those interested in such mathematical olympiads should refer to:

https://olympiads.hbcse.tifr.res.in

(I am a tutor for such mathematical olympiads).