It is quite well-known that any positive integer can be factored into a product of primes in a unique way, up to an order. (And, that 1 is neither prime nor composite) —- we all know this from our high school practice of “tree-method” of prime factorization, and related stuff like Sieve of Eratosthenes. But, it is so obvious, and so why it call it a theorem, that too “fundamental” and yet it seems it does not require a proof. It was none other than the prince of mathematicians of yore, Carl Friedrich Gauss, who had written a proof to it. It DOES require a proof — there are some counter example(s). Below is one, which I culled for my students:
(a) Show that the sum and product of elements of E are in E.
(b) Define the norm of an element by . We say that an element is prime if it is impossible to write with , and , ; we say that it is composite if it is not prime. Show that in E, 3 is a prime number and 29 is a composite number.
(c) Show that the factorization of 9 in E is not unique.
A prime number p is called a Wilson prime if . Using a computer and some programming language like C, C++, or Python find the three smallest Wilson primes.
Show that for each positive integer n equal to twice a triangular number, the corresponding expression represents an integer.
Let n be such an integer, then there exists a positive integer m such that . We then have so that we have successively
; ; and so on. It follows that
, as required.
Comment: you have to be a bit aware of properties of triangular numbers.
1001 Problems in Classical Number Theory by Jean-Marie De Koninck and Armel Mercier, AMS (American Mathematical Society), Indian Edition:
Amazon India link:
Reference: Elementary Number Theory, David M. Burton, Sixth Edition, Tata McGraw-Hill.
(We are all aware of the proof we learn in high school that is irrational. (due Pythagoras)). But, there is an interesting variation of that proof.
Let with , there must exist integers r and s such that . As a result, . This representation leads us to conclude that is an integer, an obvious impossibility. QED.
If , find the least value of .
will be minimum when will be minimum.
Now, ….call this equation I.
Hence, z will be maximum when is maximum but is the product of two factors whose sum is .
Hence, will be maximum when both these factors are equal, that is, when
. From equation I, maximum value of . Hence, the least value of .
Some basics related to maximum and minimum:
Let a and b be two positive quantities, S their sum and P their product; then, from the identity:
, we have
Hence, if S is given, P is greatest when ; and if P is given, S is least when . That is, if the sum of two positive quantities is given, their product is greatest when they are equal; and, if the product of two positive quantities is given, their sum is least when they are equal.
To find the greatest value of a product the sum of whose factors is constant.
Let there be n factors , and suppose that their sum is constant and equal to s.
Consider the product , and suppose that a and b are any two unequal factors. If we replace the two unequal factors a and b by the two equal factors , the product is increased, while the sum remains unaltered; hence, so long as the product contains two unequal factors it can be increased without altering the sum of the factors; therefore, the product is greatest when all the factors are equal. In this case, the value of each of the n factors is , and the greatest value of the product is , or .
Corollary to Basic 2:
If are unequal, ;
that is, .
By an extension of the meaning of the terms arithmetic mean and geometric mean, this result is usually stated as follows: the arithmetic mean of any number of positive quantities is greater than the geometric mean.
To find the greatest value of when is constant; m,n, p, ….being positive integers.
Solution to Basic 3:
Since m,n,p, …are constants, the expression will be greatest when is greatest. But, this last expression is the product of factors whose sum is , or , and therefor constant. Hence, will be greatest when the factors are all equal, that is, when
Thus, the greatest value is .
Some examples using the above techniques:
Show that where r is any real number.
that is, , which is the desired result.
Find the greatest value of for any real value of x numerically less than a.
The given expression is greatest when is greatest; but, the sum of the factors of this expression is , that is, ; hence, is greatest when , that is, . Thus, the greatest value is .
The determination of maximum and minimum values may often be more simply effected by the solution of a quadratic equation than by the foregoing methods. For example:
Divide an odd integer into two integral parts whose product is a maximum.
Let an odd integer be represented as ; the two parts by x and ; and the product by y; then ; hence,
but the quantity under the radical sign must be positive, and therefore y cannot be greater than , or, ; and since y is integral its greatest value must be ; in which case , or n; thus, the two parts are n and .
Sometimes we may use the following method:
Find the minimum value of .
Put ; then the expression
which in turn equals
Hence, the expression is a a minimum when the square term is zero; that is when .
Thus, the minimum value is , and the corresponding value of x is .
Problems for Practice:
- Find the greatest value of x in order that may be greater than .
- Find the minimum value of , and the maximum value of .
- Show that and .
- Find the maximum value of when x lies between 7 and -2.
- Find the minimum value of .
If , , be the roots of the cubic equation . Prove that the equation in y whose roots are is obtained by the transformation . Hence, form the equation with above roots.
Given that , , are the roots of the equation:
…call this equation I.
By relationships between roots and co-efficients, (Viete’s relations), we get
and , and
Now, , that is,
, or …call this equation II.
Subtracting Equation II from Equation I, we get
which is the required transformation.
Now, , that is,
Putting this value of x in Equation I, we get
, that is,
, which is the required equation.