# Some Number Theory Questions for RMO and INMO

1) Let $n \geq 2$ and k be any positive integers. Prove that $(n-1)^{2}\mid (n^{k}-1)$ if and only if $(n-1) \mid k$.

2) Prove that there are no positive integers a, b, $n >1$ such that $(a^{n}-b^{n}) \mid (a^{n}+b^{n})$.

3) If a and $b>2$ are any positive integers, prove that $2^{a}+1$ is not divisible by $2^{b}-1$.

4) The integers 1,3,6,10, $\ldots$, $n(n+1)/2$, …are called the triangular numbers because they are the numbers of dots needed to make successive triangular arrays of dots. For example, the number 10 can be perceived as the number of acrobats in a human triangle, 4 in a row at the bottom, 3 at the next level, then 2, then 1 at the top. The square numbers are $1, 4, 9, \ldots, n^{2}, \ldots$ The pentagonal numbers 1, 5, 12, 22, $\ldots$, $(3n^{2}-n)/2$, $\ldots$, can be seen in a geometric array in the following way: Start with n equally spaced dots $P_{1}, P_{2}, \ldots, P_{n}$ on a straight line in a plane, with distance 1 between consecutive dots. Using $P_{1}P_{2}$ as a base side, draw a regular pentagon in the plane. Similarly, draw $n-2$ additional regular pentagons on base sides $P_{1}P_{3}$, $P_{1}P_{4}$, $\ldots$, $P_{1}P_{n}$, all pentagons lying on the same side of the line $P_{1}P_{n}$. Mark dots at each vertex and at unit intervals along the sides of these pentagons. Prove that the total number of dots in the array is $(3n^{2}-n)/2$. In general, if regular k-gons are constructed on the sides $P_{1}P_{2}$, $P_{1}P_{3}$, …, $P_{1}P_{n}$, with dots marked again at unit intervals, prove that the total number of dots is $1+kn(n-1)/2 -(n-1)^{2}$. This is the nth k-gonal number.

5) Prove that if $m>n$, then $a^{2^{n}}+1$ is a divisor of $a^{2^{m}}-1$. Show that if a, m, n are positive with $m \neq n$, then

$( a^{2^{m}}+1, a^{2^{n}}+1) = 1$, if a is even; and is 2, if a is odd.

6) Show that if $(a,b)=1$ then $(a+b, a^{2}-ab+b^{2})=1$ or 3.

7) Show that if $(a,b)=1$ and p is an odd prime, then $( a+b, \frac{a^{p}+b^{p}}{a+b})=p$ or 1.

8) Suppose that $2^{n}+1=xy$, where x and y are integers greater than 1 and $n>0$. Show that $2^{a}\mid (x-1)$ if and only if $2^{a}\mid (y-1)$.

9) Prove that $(n!+1, (n+1)!+1)=1$.

10) Let a and b be positive integers such that $(1+ab) \mid (a^{2}+b^{2})$. Show that the integer $(a^{2}+b^{2})/(1+ab)$ must be a perfect square.

Note that in the above questions, in general, (a,b) means the gcd of a and b.

More later,
Nalin Pithwa.

# A Primer: Generating Functions: Part II: for RMO/INMO 2019

We shall now complicate the situation a little bit. Let us ask for the combinations of the symbols $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ with repetitions of each symbol allowed once more in the combinations. For example, let there be only two symbols $\alpha_{1}, \alpha_{2}$. Let us look for combinations of the form:

$\alpha_{1}$, $\alpha_{2}$, $\alpha_{1}\alpha_{2}$, $\alpha_{1}\alpha_{1}$, $\alpha_{2}\alpha_{2}$, $\alpha_{1}\alpha_{1}\alpha_{2}$, $\alpha_{1}\alpha_{2}\alpha_{2}$, $\alpha_{1}\alpha_{1}\alpha_{2}\alpha_{2}$

where, in each combination, each symbol may occur once, twice, or not at all. The OGF for this can be constructed by reasoning as follows: the choices for $\alpha_{1}$ are not-$\alpha_{1}$, $\alpha_{1}$ once, $\alpha_{1}$ twice. This is represented by the factor $(1+\alpha_{1}t+\alpha_{1}^{2}t^{2})$. Similarly, the possible choices for $\alpha_{2}$ correspond to the factor $(1+\alpha_{2}t+\alpha_{2}^{2}t^{2})$. So, the required OGF is $(1+\alpha_{1}t+\alpha_{1}^{2}t)(1+\alpha_{2}t+\alpha_{2}^{2}t^{2})$

On expansion, this gives : $1+(\alpha_{1}+\alpha_{2})t+(\alpha_{1}\alpha_{2}+\alpha_{1}^{2}+\alpha_{2}^{2})t^{2}+(\alpha_{1}^{2}\alpha_{2}+\alpha_{1}\alpha_{2}^{2})t^{3}+(\alpha_{1}^{2}\alpha_{2}^{2})t^{4}$

Note that if we omit the term 1 (which corresponds to not choosing any $\alpha$), the other 8 terms correspond to the 8 different combinations listed in (*). Also, observe that the exponent r of the $t^{r}$ tells us that the coefficient of $t^{r}$ has the list or inventory of the r-combinations (under the required specification — in this case, with the restriction on repetitions of symbols) in it:

$\bf{Illustration}$

In the light of the foregoing discussion, let us now take up the following question again: in how many ways, can a total of 16 be obtained by rolling 4 dice once?; the contribution of each die to the total is either a “1” or a “2” or a “3” or a “4” or a “5” or a “6”. The contributions from each of the 4 dice have to be added to get the total — in this case, 16. So, if we write: $t^{1}+t^{2}+t^{3}+t^{4}+t^{5}+t^{6}$

as the factor corresponding to the first die, the factors corresponding to the other three dice are exactly the same. The product of these factors would be:

(*) $(t+t^{2}+t^{3}+t^{4}+t^{5}+t^{6})^{4}$

Each term in the expansion of this would be a power of t, and the exponent k of such a term $t^{k}$ is nothing but the total of the four contributions which went into it. The number of times a term $t^{k}$ can be obtained is exactly the number of times k can be obtained as a total on a throw of the four dice. So, if $\alpha_{k}$ is the coefficient of $t^{k}$ in the expansion, $\alpha_{16}$ is the answer for the above question. Further, since (*) simplifies to $(\frac{t(1-t^{6})}{1-t})^{4}$, it follows that the answer for the above question tallies with the coefficient specified in the following next question: calculate the coefficient of $t^{12}$ in $(\frac{(1-t^{6})}{(1-t)})^{4}$.6

Now, consider the following problem: Express the number $N(n,p)$ of ways of obtaining a total of n by rolling p dice, as a certain coefficient in a suitable product of binomial expansions in powers of t. [ this in turn, is related to the observation that the number of ways a total of 16 can be obtained by rolling 4 dice once is the same as the coefficient of $t^{12}$ in $(\frac{1-t^{6}}{1-t})^{4}$]:

So, we get that $N(n,p)=$ coefficient of $t^{n-p}$ in $(\frac{1-t^{6}}{1-t})^{p}$

Let us take an example from a graphical enumeration:

A $\it {graph}$ $G=G(V,F)$ is a set V of vertices a, b, c, …, together with a set $E=V \times V$ of $\it {edges}$ $(a,b), (a,a), (b,a), (c,b), \ldots$ If $(x,y)$ is considered the same as $(y,x)$, we say the graph is $\it{undirected}$. Otherwise, the graph is said to be $\it{directed}$, and we say ‘$(a,b)$ has a direction from a to b’. The edge $(x,x)$ is called a loop. The graph is said to be of order $|V|$.

If the edge-set E is allowed to be a multiset, that is, if an edge $(a,b)$ is allowed to occur more than once, (and, this may be called a ‘multiple edge’), we refer to the graph as a general graph.

If $\phi_{5}(n)$ and $\psi_{5}(n)$ denote the numbers of undirected (respectively, directed) loopless graphs of order 5, with n edges, none of them a multiple edge, find the series $\sum \phi_{5}(n)t^{n}$ and $\sum \psi_{5}(n)t^{n}$.

Applying our recently developed techniques to the above question, a graph of 5 specified vertices is uniquely determined once you specify which pairs of vertices are ‘joined’. Suppose we are required to consider only graphs with 4 edges. This would need four pairs of vertices to be selected out of the total of $5 \choose 2$ equal to 10 pairs that are available. So selection of pairs of vertices could be made in $10 \choose 4$ ways. Each such selection corresponds to one unique graph, with the selected pairs being considered as edges. More informally, having selected a certain pairs of vertices, imagine that the vertices are represented by dots in a diagram and join the vertices of each selected pair by a running line. Then, the “graph” becomes a “visible” object. Note that the number of graphs is just the number of selections of pairs of vertices. Hence, $\phi_{5}(4)=10 \choose 4$.

Or, one could approach this problem in a different way. Imagine that you have a complete graph on 5 vertices — the “completeness” here means that every possible pair of vertices has been joined by an edge. From the complete graph which has 10 edges, one has to choose 4 edges — any four, for that matter — in order to get a graph as required by the problem.

On the same lines for a directed graph, one has a universe of 10 by 2, that is, 29 edges to choose from, for, each pair x,y gives rise to two possible edges $(x,y)$ and $(y,x)$. Hence,

$\psi_{5}(4)=20 \choose 4$.

Thus, the counting series for labelled graphs on 5 vertices is $1 + \sum_{p=1}^{10} {10 \choose p}t^{p}$
and the counting series for directed labelled graphs on 5 vertices is
$1+ \sum_{p=1}^{20}{20 \choose p}t^{p}$.

Finally, the OGF for increasing words on an alphabet ${a,b,c,d,e}$ with $a is

$(1+at+a^{2}t^{2}+\ldots)(1+bt+b^{2}t^{2}+\ldots)(1+ct+c^{2}t^{2}+\ldots)\times (1+dt+d^{2}t^{2}+\ldots)(1+et+e^{2}t^{2}+\ldots)$

The corresponding OE is $(1+t+t^{2}+t^{3}+\ldots)^{5}$ which is nothing but $(1-t)^{-5}$ (this explains the following problem: Verify that the number of increasing words of length 10 out of the alphabet $\{a,b,c,d,e \}$ with $a is the coefficient of $t^{10}$ in $(1-t)^{-5}$ ).

We will continue this detailed discussion/exploration in the next article.

Until then aufwiedersehen,
Nalin Pithwa

# Elementary problems in Ramsey number theory for RMO

Question 1:

Show that in any group of 6 people there will always be a subgroup of 3 people who are pairwise acquainted or a subgroup of 3 people who are pairwise strangers.

Solution 1:

Let $\{ A, B, C, D, E, F\}$ be a group of 6 people. Suppose that the people known to A are seated in room Y and the people NOT known to A are seated in room Z; A is not in either room. Then, there are necessarily at least 3 people in either room Y or in room Z; (a) Suppose B, C, D to be in room Y. Either these 3 people are mutual strangers (and so the given theorem is true), or, at least two of them (say, B and C) know each other. In the latter case, A, B and C form a group of 3 mutual acquaintances — and again, the theorem is true. (b) In (a), replace room Y by Z and interchange the notion of ‘”acquaintances” and “strangers”‘.

Question 2:

Show that in any group of 10 people there is always (a) a subgroup of 3 mutual strangers or a subgroup of 4 mutual acquaintances, and (b) a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 2:

(a) Let A be one of the ten people; the remaining 9 people can be assigned to two rooms: those who are known to A are in room Y and those who are not known to A are in room Z. Either room Y has at least 6 people or room Z has at least 6 people. For, (i) suppose room Y has at least 6 people. Then, by previous problem number 1, there is either a subgroup of 3 mutual acquaintances or a subgroup of 3 mutual strangers (thus, the theorem is true) in this room. In the former case, A and these 3 people constitute 4 mutual acquaintances (ii) Suppose room Z has at least 4 people. Either these 4 people know one another or at least 2 of them, say B and C, do not know each other. In the former case, we have a subgroup of 4 mutual acquaintances. In the latter case A, B and C constitute 3 mutual strangers.

(b) In the previous scenario, let people who are strangers become acquaintances, and let people who are acquaintances pretend they are strangers. The situation is symmetric.

Question 3:

Show that in any subgroup of 20 people there will always be either a subgroup of 4 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 3:

Suppose A is one of these 20 people. People known to A are in room Y and people not known to A are room Z. Either room Y has at least 10 people or room Z has at least 10 people. (i) If Y has at least 10 people, then by part B of problem number 2 here, there is either a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers — as asserted — in this room. In the former case, A and these mutual acquaintances will form a subgroup of 4 mutual acquaintances. (ii) Switch ‘”acquaintances” and “strangers”‘ in (i).

Question 4:

Let p and q be 2 positive integers. A positive integer r is said to have the (p,q) – Ramsey property, if in any group of r people either there is a subgroup of p people known to one another or there is a subgroup of q people not known to one another. {By Ramsey’s theorem, all sufficiently large integers r have the (p,q)-Ramsey property.} The smallest r with the (p,q)-Ramsey property is called the Ramsey number R(p,q). Show that (a) R(p,q) = R (q,p). (b) $R(p,1)=1$, and (c) R(p,2)=p.

Solution 4:

(a) By parts (b) of the previous three questions, we have proved part a of the proof here.

(b) This is obvious.

(c) In any group of p people, if all of them are not known to one another, there will be at least 2 people who do not know each other.

Question 5:

Prove that $R(3,3)=6$.

Solution 5:

Question 1 and its proof in this blog article imply that $R(3,3) \leq 6$.

To prove that $R(3,3)>5$, it is sufficient to consider a seating arrangement of 5 people about a round table in which each person knows only the 2 people on either side. In such a situation, there is no set of 3 mutual acquaintances and no set of 3 people not known to one another.

Question 6:

Show that if m and n are integers both greater than 2, then

$R(m,n) \leq R(m-1,n) + R(m,n-1)$.

(this recursive inequality gives a non-sharp upper bound for R(m,n)).

Solution 6:

Let $p \equiv R(m-1,n)$, $q=R(m,n-1)$ and $r \equiv p + q$. Consider a group $\{ 1,2, 3, \ldots, r\}$ of r people. Let L be the set of people known to person 1 and M be the set of people NOT known to person 1. The two sets together have $r-1$ people, so either L has at least p people or M has at least q people. (a) If L has $p \equiv R(m-1,n)$ people, then, by definition, it contains a subset of $(m-1)$ people known to one another or it contains a subset of n people unknown to one another. In the former case, the $(m-1)$ people and person 1 constitute m people known to one another.

Thus, in their case, a group of $R(m-1,n) + R(m,n-1)$ people necessarily includes m mutual acquaintances or n mutual strangers. That is, $R(m,n) \leq R(m-1,n) + R(m,n-1)$.

(b) By the usual symmetry argument, the same conclusion follows when M contains q people.

Question 7:

(Remark: A pretty property of Ramsey numbers related to combinatorics).

Show that if m and n are integers greater than 1, then $R(m,n) \leq { {m+n-2} \choose {m-1}}$ — a non-recursive upper bound.

Solution 7:

When $m=2$, or $n=2$, (i) holds with equality (see problem 4 in this blog article). The proof is by induction on $k=m+n$. As we have just seen, the result is true when $k=4$. Assume the result true for $k-1$. Then,

$R(m-1,n) \leq {{m+n-3} \choose {m-2}}$  and $R(m,n-1) \leq {{m+n-3} \choose {m-1}}$

Now, Pascal’s identity gives:

${{m+n-3} \choose {m-2}} + {{m+n-3} \choose {m-1}} = {{m+n-2}} \choose {m-1}$ so that $R(m-1,n) + R(m,n-1) \leq {{m+n-2}} \choose {m-1}$

But, from the previous question and its solution, we get $R(m,n) \leq R(m-1,n) + R(m, n-1)$

PS: As Richard Feynman, used to say, you will have to “piddle” with smallish problems as particular cases of these questions in order to get a grip over theory or formal language of this introduction.

PS: Additionally, you can refer to any basic Combinatorics text like Brualdi, or Alan Tucker or even Schaum Series outline ( V K Balakrishnan).

# Some number theory problems: tutorial set II: RMO and INMO

1. A simplified form of Fermat’s theorem: If x, y, z, n are natural numbers, and $n \geq z$, prove that the relation $x^{n} + y^{n} = z^{n}$ does not hold.

2. Distribution of numbers: Find ten numbers $x_{1}, x_{2}, \ldots, x_{10}$ such that (a) the number $x_{1}$ is contained in the closed interval $[0,1]$ (b) the numbers $x_{1}$ and $x_{2}$ lie in different halves of the closed interval $[0,1]$ (c) the numbers $x_{1}$, $x_{2}$, $x_{3}$ lie in different thirds of the closed interval $[0,1]$ (d) the numbers $x_{1}$, $x_{2}$, $x_{3}$ and $x_{4}$ lie in different quarters of the closed interval $[0,1]$,  etc., and finally, (e) the numbers $x_{1}$, $x_{2}$, $x_{3}, \ldots, x_{10}$ lie in different tenths of the closed interval $[0,1]$

3. Is generalization of the above possible?

4. Proportions: Consider numbers A, B, C, p, q, r such that: $A:B =p$, $B:C=q$, $C:A=r$, write the proportion $A:B:C = \Box : \Box : \Box$ in such a way that in the empty squares, there will appear expressions containing p, q, r only; these expressions being obtained by cyclic permutation of one another expressions.

5. Give an elementary proof of the fact that the positive root of $x^{5} + x = 10$ is irrational.

I will give you sufficient time to try these. Later, I will post the solutions.

Cheers,

Nalin Pithwa.

# Some number theory training questions: RMO and INMO

Question 1:

Let us write an arbitrary natural number (for example, 2583), and then add the squares of its digits. ($2^{2}+5^{2}+8^{2}+3^{2}=102$). Next, we do the same thing to the number obtained. Namely, $1^{2}+0^{2}+2^{2}=5$. Now proceed further in the same way:

$5^{2}=25$, $2^{2}+5^{2}=29$, $2^{2}+9^{2}=85, \ldots$.

Prove that unless this procedure leads to number 1 (in which case, the number 1 will, of course, recur indefinitely), it must lead to the number 145, and the following cycle will repeat again and again:

145, 42, 20, 4, 16, 37, 58, 89.

Question 2:

Prove that the number $5^{5k+1} + 4^{5k+2} + 3^{5k}$ is divisible by 11 for every natural k.

Question 3:

The number $3^{105} + 4^{105}$ is divisible by 13, 49, 181 and 379, and is not divisible by either 5 or by 11. How can this result be confirmed?

Cheers,

Nalin Pithwa.