The Pedagogical Aspects of Teaching Prime Numbers to
Gifted Children
– A classroom Experience.
Author: Dr. S.R. SANTHANAM.
Email: santhanam_sr04@yahoo.com
Abstract
This paper is an outcome of the classroom experience of the author dealing with
gifted children. The concepts and problem solving techniques on primes are
introduced to gifted children of age groups 10-20 in Indian schools. Analysis and
tackling the problems by the children were carefully noted. The interesting
aspects of this experience are shared in this paper.
Concept of a Prime Number
The concept of prime numbers is introduced to students of age group 11-12 in
schools. This concept is closely connected to divisors of a number. From around
2500 years onwards, prime numbers posed challenges to even a mathematician
of high order. Till now nobody knows how prime numbers are sprinkled on the
number line. So there is no definite expression either algebraic or any other to
give all the primes. This makes the problems on primes difficult. But with no
definite formula, no deep information about the primes, lot of problems on prime
numbers can be given to gifted children. These problems involve simple
mathematical techniques with logical reasoning. This short paper is about a
series of interaction in the class room of children of different age groups 11-15.
The informal way of introducing a prime may be as follows.
Some positive integers can be written as product of two or more strictly smaller
positive integers. When we give this idea and give a composite number to a
student, he or she decomposes it as a product of two or more positive integers.
The number 12 was given to a set of students of age group 11 – 12. Most of them
wrote 12 = 34 or 43. Only few of them wrote 12=322. It is a surprise that
nobody wrote 12 = 3221. The number 2 repeating itself twice is allowed or not
was a doubt for many children. After the clarification that the numbers in the
decomposition can repeat, then almost all of them, without knowing the theorem
on prime decomposition of an integer, could manage.
It is quite an interesting observation when the number 3 was given. Most of them
reported that it cannot be written as a product of two smaller integers. Only very
few observed that 3=13, is the way.
Giving several composite and primes to them, they realized that some numbers
can be written as a product of two or more than two positive integers and some
can only be written as a product of 1 and that number only. Now it is an easy task
for the teacher to give specific names to the numbers as prime or composite.
The following concepts were then introduced.
The only even prime is 2, and all the other primes are odd.
All odd numbers need not be primes.
A simple question from the side of a child of age group 11 – 12 will certainly
embarrass the teacher.
How many primes are there totally?
It is very difficult on the part of a teacher to tell a definite answer because the
concept of infinity comes and it is difficult to make a11 year old child no
understand the concept.
The above mentioned concepts are the only concepts regarding a prime number
which come through all the classes of a school, to children of age group 15 – 16.
With these few concepts the job of a teacher becomes difficult to frame
questions. But surprisingly there are innumerable problems on primes and almost
all of them are of non-routine type. This is because there is no specific formula for
generating all primes.
A small hurdle on the part of the teacher is to convince them that 1 cannot be
considered as a prime.
Questions on Prime Numbers to a primary child
What kind of questions can be given to a primary school child on primes? The
following is one such.
Prove that there are no five digit numbers whose digits are different and primes.
When this was given to a set of primary school children (age group 8 -10) many
of them gave the solution. They wrote first few primes, 2, 3, 5, 7, 11, 13… They
started framing 5-digit numbers of different digits (i.e) each digit must be a single
digit prime. They form 2357, a 4-digit number. For the 5th digit there is no single
digit prime different from 2, 3, 5 and 7.
Prime Number to a middle school child
When a student of age group 13 – 15 comes across the statement, there is no
algebraic polynomial which gives all the primes, of course the proof being out of
capacity, always tries to get an algebraic polynomial. Even the giant of a
mathematician Karl Frederick Gauss gave a polynomial x2+x+41 which gives
primes for many consecutive values of x (0 to 40) realizing that for x=41, the
formula fails). Many great mathematicians tried and failed. But the trials
culminated in many remarkable results.
It is a common outcome on the part of a child, when prime numbers are known,
why don’t a formula can be found? Many students of age group 13 – 15 try and
get linear expressions of the type 4n-1,4n+3,4n+1,6n+1,6n+5,6n-1 etc which give
fairly a good amount of primes.
For example, let us take 4n-1
Put n=1, 2, 3, we get 3, 7, 11,
But fails to give a prime for n=4.
Each one of the expressions gives several primes (Purposely the word infinity is
avoided). The child of age group 13 – 15 is convinced about this fact.
Now the teacher without mentioning the infinitude of primes can give several
problems to the children which involve only logic.
Non-routine problem on primes
The following problem is one such.
Can a prime number p be found such that 2p+1 and 4p+1 are also primes?
This problem cannot be solved in a routine way because there is no formula for a
prime. So, first of all the problem can be analyzed by trying few values of primes.
When p=2, 3p+1=5, 4p+1=9. Here 2p+1 is a prime but 4p+1 is not a prime.
When p=3, 2p+1=7, 4p+1=13. Here 2p+1 and 4p+1 are both primes.
So there exists a p, for which 2p+1 and 4p+1 are both primes. But that is not the
method to find p.
Another random check, for example, p=5, 2p+1 is prime but 4p+1 is not.
Now the student can be asked to check for any prime greater than 3.
Any number when divided by 3 leaves remainder 0, 1 or 2, therefore of the form
3k+1 or 3k+2.
If p=3k+1 then 2p+1 = 2(3k+1)+1 = 6k+3=3(2k+1) which is divisible by 3, then not
a prime.
If p=3k+2, then 4p+1 =4(3k+2)+1=12k+9=3(4k+3) which is divisible by 3, then not
a prime.
Thus for any prime p>3, either 2p+1 or 4p+1 is composite. So the conclusion is
the only prime p for which 2p+1 and 4p+1 are primes is 3.
The problems on primes, even though they are not very tough, employ different
logic to solve. They are totally non-routine in nature. Of course, there is no
formula for a prime number, no algebraic polynomial of any degree which
generates only primes, but some forms of a prime numbers will be very useful in
solving problems.
Students of age group 12 – 15 are given the following results.
Every odd prime can be put either in the form 4k+1 or 4k+3 (or 4k-1) where k is a
positive integer.
When this result is given to the children of age group 12-15, there is every
possibility that a child thinks that 4k+1 or 4k-1 is a formula for a prime. It is the
task of the teacher to clarify the important difference between every odd prime
can be put in the form 4k+1 or 4k-1 but 4k+1 or 4k-1 will not give always primes.
17 is a prime and 17=44+1, if k=4, 4k+1 is equal to 17, but if k=5, 4k+1 is 21
which is not a prime.
This important and subtle point is to be properly driven into the minds of children.
What kind of a proof is to be given that every odd prime can be put either in the
form 4k+1 or 4k+3, where k is a positive integer?
For an odd prime n, when divided by 4 leaves a remainder 1or 2 or 3 (remainder
0 means n=4k n is composite).
 n=4k+1 or 4k+2 or 4k+3. Now 4k+2 =2(2k+1) is composite
 b=4k+1 or 4k+3 (or 4k-1)
Similarly we can prove every odd prime greater than 3, is of the form 6n+1 or
6n+5 for positive integers n.
The following problem is given to gifted children of age group 12 – 13
If p is a prime greater than 3, then p2 -1 is divisible by 24.
As expected they started giving prime values for p and checking whether p2 -1 is
divisible by 24 or not.
p = 5 gives p2-1 =24, p =7 gives p2-1 =48, p = 11 gives p2-1=120, and all of
them are divisible by 24.
They are convinced about the result. Now they started trying for a proof. Again as
expected p2-1 was factorized as (p-1) (p+1).
When p is greater than 3 and prime; it must be an odd number so that p-1 and
p+1 are both even and each greater than 2. So they concluded that p2-1is
divisible by 8. The remaining part is to prove p2-1is also divisible by 3. This posed
some difficulty to them. But some of them could manage in the following way.
When the consecutive natural numbers are taken then one of them is a multiple
of 3. Now (p – 1) and (p +1) are not consecutive. But (p -1), p, (p +1) are
consecutive. Since p > 3 and prime so not divisible by 3, which implies that either
p-1 or p+1 is divisible by 3. Thus p2-1 is divisible by 24.
A teacher finds that each problem on primes needs different strategies.
The following problem was given to a set of gifted children
If p and p2+2 are primes, will p3+2 a prime?
A lot of interaction was going among them in groups.
Almost all students concluded that p must be odd for when p=2, p2+2=6 not a
prime.
They tried for p=3. p2+2=9+2=11is a prime.
The statement is of course true.
 p3+2=27+2=29 is a prime
But if an additional question is asked, as how many such primes p exists?
It is to be verified that for what prime p, p2+2 is also a prime.
For p=5, p2+2=27 not a prime, p=7, p2+2=51 not a prime, p=11, p2+2=121+2=123
not a prime
A gifted child suspects that the only p for which p2+2 and p3+2 are primes is 3.
An odd prime can be one of the several forms 4n  1, 6n  1.
If p=4n1 then p2+2=(4n1)2+2=16n28n+3
 It is a multiple by 3, therefore composite.
If p=6n1 then p2+2 =(6n1)2+2=36n212n+3
 Any odd prime greater than 5 will not yield p2+2 a prime 3.
The following problem kindled the interest of gifted children very much.
Find all positive integers n, such that each of the numbers n+1, n+3, n+7, n+9,
n+13 and n+15 is a prime.
A lot of interaction was going among them in groups. They all started trying
putting n=1,2,3…
n=1 is immediately rejected. n=2 is rejected because n+7 is composite, n=3 is
rejected because of n+1.
n=4 is accepted with a joy among them, since they got 5, 7, 11, 13, 17, 19 and all
of them are primes.
n=5 is rejected because n+1 is composite.
They realized that any odd n makes n+1 and others composite so even n only are
the eligible candidates.
n=6 gives n+9 divisible by 5, n=8 gives n+7 composite divisible by 5, n=10 gives
n+15 divisible by 5 and n=12 gives n+3 divisible by 5
Thus they found for any even n, one of the numbers in the sequence is divisible
by 5.
But how to prove it?
The catch is the numbers 1, 3, 7, 9, 13, 15
The remainders got when these numbers divided by 5 are 1, 3, 2, 4, 3, 0
respectively.
Therefore n+1, n+3, n+7, n+9, n+13 and n+15 give also all possible remainders
when divided by 5. Thus atleast one of them is divisible by 5.
A problem with slightly different technique is to be employed was given to a set of
gifted children of age group 13-15.
Find all primes p such that p+2 and p2+2p-8 are also primes.
When this was given, the students, as usual started giving various primes values
for p.
p=3  p+2=5 and p2+2p-8=9+6-8=7
 p=3 is a value sought for.
p=5  p+2=7and p2+2p-8=25+10-8=27
Here p2+2p-8 is not a prime.
Now the students tried for p=7 where p+2 is not a prime, p=9, p2+2p-8 is not a
prime, p=11 p2+2p-8=125 is not a prime.
Then they got suspicion that p2+2p-8 may not be a prime when p+2 is a prime for
any p>3. So p2+2p-8 gets the full focus. Some of them immediately factorized this
quadratic expression as (p-2)(p+4). If it is a prime, one factor must be 1.
 p-2=1 p=3
p+4=1  p=-3 is not permissible.
 p=3 is the only value.
This factorization method is a very useful method in problem involving primes.
At this juncture the teacher proposed a challenge. Can any student frame a
question similar to the one solved.
Some students framed questions resembling the previous problem. Few of them
are given below.
Find all primes p such that p+4 and p2+6p-16 are primes.
Find primes p such that p+2 and p2+2p-24 are primes.
Find all primes p such that p+5 and p2+20p-21 are primes.
Since the students learnt the trick of framing such question, the teacher went
further deep. The following question was posed to them.
Can you frame a question involving cubic expression which gives primes for one
or more values of the variable involved?
The teacher facilitated them with the following question. Find the all primes p
such that p3+3p2-5p-10 is also a prime. Since the students realized that the
factorization of the quadratic gave way to such problem, it was not a surprise that
when the above problem was given, the children started factorizing the cubic.
Let f(p)= p3+3p2-5p-10, f(1)=1+3-5-100, f(2)=8+12-10-10=0
 (p-2) is a factor. The other factors can be got by synthetic division.
2
1
3  5  10
0
2 10
10
1
5
0
5
3
2
2
 p +3p -5p-10 = (p-2) (p +5p+5)
If this is a prime, these are two possibilities.
p-2=1 and p2+5p+5 is a prime or p2+5p+5=1 and p-2 is a prime
Case (1) when p-2=1 then p=3 and p2+5p+5 =9+15+5=29 which is a prime.
Case (2) p2+5p+5 =1  p2+5p+4=0  there exist no p (since p is a positive
integer)
Thus the only solution is p = 3.
References:
Arthur Engel – Problem solving strategies – Springer International Edition –
First Indian Reprint – 2003.
Dmitri Fomin et all - Mathematical circles ( Russian Experience – University
Press).
Waclaw Sierpinski – 250 problems in Elementary Number Theory – Elsevier.