Warm up

MOPSS
8 May 2025

Mathematics Olympiad
Problem Solving Sessions

MOPSS

Department of Mathematics
IISER Bhopal

https://jpsaha.github.io/MOTP/MOPSS/

Suggested readings
• Evan Chen’s advice On reading solutions, available at https://blog.
evanchen.cc/2017/03/06/on-reading-solutions/.
• Evan Chen’s Advice for writing proofs/Remarks on English, available at
https://web.evanchen.cc/handouts/english/english.pdf.
• Notes on proofs by Evan Chen from OTIS Excerpts [Che25, Chapter 1].
• Tips for writing up solutions by Edward Barbeau, available at https:
//www.math.utoronto.ca/barbeau/writingup.pdf.
• Evan Chen discusses why math olympiads are a valuable experience for
high schoolers in the post on Lessons from math olympiads, available at
https://blog.evanchen.cc/2018/01/05/lessons-from-math-olympiads/.
List of problems and examples
1.1 Example (G. Galperin, Tournament of Towns, Autumn 1989,
Junior, O Level, P4) . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Example (IMO 1959 P1, proposed by Poland) . . . . . . . . 2
1.3 Example (India RMO 2019a P4) . . . . . . . . . . . . . . . . 3
1.4 Example (Tournament of Towns, Spring 2014, Senior, A Level,
P4 by I. I. Bogdanov) . . . . . . . . . . . . . . . . . . . . . . 4
1.5 Example (Tournament of Towns, Spring 2020, Junior, O Level,
P4 by Alexandr Yuran) . . . . . . . . . . . . . . . . . . . . . 4
1.6 Example (India RMO 2015e P3) . . . . . . . . . . . . . . . . 4
1.7 Example (India RMO 2016e P5) . . . . . . . . . . . . . . . . 6

§1 Warm up
It would be good to go through [Che25, Chapter 1, Notes on proofs].
Example 1.1 (G. Galperin, Tournament of Towns, Autumn 1989, Junior, O
Level, P4). Find the solutions of the equation
1 10
x+ 1 = (1)
y+ z
7
in positive integers.

Solution 1. Let x, y, z be positive integers satisfying Eq. (1). Since y ≥ 1 and

1 1 1
z > 0, it follows that y + z > 1, which gives 0 < y+ 1 < 1. Using Eq. (1), it
z
follows that x = 1, and hence y + z1 = 73 . By a similar argument as above1 , it
follows that y = 2 and consequently, z = 3.
Moreover, for x = 1, y = 2, z = 3, Eq. (1) holds.
This proves that x = 1, y = 2, z = 3 is the only solution2 of Eq. (1). ■

Remark. Is the statement that for x = 1, y = 2, z = 3, Eq. (1) holds in the

above argument redundant? Or, is it not so? Think about it. Further, it would
be worth going through [Che25, Chapter 1].

Example 1.2 (IMO 1959 P1, proposed by Poland). Prove that the fraction
21n + 4
14n + 3
is irreducible for every natural number n.
1Write the argument instead of resorting to using “by a similar argument” unless it is clear
to you. Even then, consider it as an exercise and write it down!
2 It means x = 1, y = 2, z = 3 is a solution to Eq. (1), and that it is the only solution, i.e. if

we are given a solution, it cannot be different from x = 1, y = 2, z = 3. Does the above

argument prove both?

2
1 Warm up Typos may be reported to jpsaha@iiserb.ac.in.

We need to show that 21n + 4, 14n + 3 have no factor in common other than
1 for every natural number n.

Summary — It follows from considering the greatest common divisor of the

numerator and the denominator.

Walkthrough —
• The summand 21n from the numerator and the summand 14n from the
denominator do not “balance well”.
• One way “enforce balancing” would be to consider

2 · 21n − 3 · 14n,

which vanishes.
• Does the above “ad hoc thoughts” help to conclude?

Solution 2. Let n be a natural number. It is enough to show that the greatest

common divisor of the integers 21n + 4, 14n + 3 is equal to 1. Note that any
common divisor of 21n + 4, 14n + 3 divides
2(21n + 4) − 3(14n + 3) = −1.
This shows that the greatest common divisor of the integers 21n + 4, 14n + 3 is
equal to 1, completing the proof. ■

Example 1.3 (India RMO 2019a P4). Consider the following 3 × 2 array
formed by using the numbers 1, 2, 3, 4, 5, 6,
   
a11 a12 1 6
a21 a22  = 2 5 .
a31 a32 3 4
Observe that all row sums are equal, but the sum of the square of the squares
is not the same for each row. Extend the above array to a 3 × k array (aij )3×k
for a suitable k, adding more columns, using the numbers 7, 8, 9, . . . , 3k such
that
k
X k
X k
X k
X k
X k
X
a1j = a2j = a3j and (a1j )2 = (a2j )2 = (a3j )2
j=1 j=1 j=1 j=1 j=1 j=1

Solution 3. Note that

 
1 6 2+6 5+6 3+2·6 4+2·6
2 5 3 + 6 4 + 6 1 + 2 · 6 6 + 2 · 6
3 4 1+6 6+6 2+2·6 5+2·6
works. ■

Some style files, prepared by Evan Chen, have been adapted here. 3
8 May 2025 https://jpsaha.github.io/MOTP/

Example 1.4 (Tournament of Towns, Spring 2014, Senior, A Level, P4 by I. I.

Bogdanov). The King called two wizards. He ordered First Wizard to write
down 100 positive real numbers (not necessarily distinct) on cards without
revealing them to Second Wizard. Second Wizard must correctly determine all
these numbers, otherwise both wizards will lose their heads. First Wizard is
allowed to provide Second Wizard with a list of distinct numbers, each of which
is either one of the numbers on the cards or a sum of some of these numbers.
He is not allowed to tell which numbers are on the cards and which numbers
are their sums. Finally the King tears as many hairs from each wizard’s beard
as the number of numbers in the list given to Second Wizard. What is the
minimal number of hairs each wizard should lose to stay alive?
Example 1.5 (Tournament of Towns, Spring 2020, Junior, O Level, P4 by
Alexandr Yuran). For some integer n, the equation x2 +y 2 +z 2 −xy−yz−zx = n
has an integer solution x, y, z. Prove that the equation x2 + y 2 − xy = n also
has an integer solution x, y.

Walkthrough — Note that the identity

x2 + y 2 + z 2 − xy − yz − zx = (x − y)2 − (x − y)(z − y) + (z − y)2

holds.

Example 1.6 (India RMO 2015e P3). Find all fractions which can be written
simultaneously in the forms
7k − 5 6ℓ − 1
and
5k − 3 4ℓ − 3
for some integers k, ℓ.

Solution 4. The solution relies on the following claim.

Claim — Suppose k, ℓ are integers. Then the equality

7k − 5 6ℓ − 1
=
5k − 3 4ℓ − 3
is equivalent to the pair (k, ℓ) being equal to one of

(0, 6), (1, −1), (6, −6), (13, −7), (−2, −22), (−3, −15), (−8, −10), (−15, −9).
(2)

Proof of the Claim. Suppose k, ℓ are integers. Observing that 5k − 3 and 4ℓ − 3

are nonzero, it follows that
7k − 5 6ℓ − 1
=
5k − 3 4ℓ − 3

4 The content posted here and at this blog by Evan Chen are quite useful.
1 Warm up Typos may be reported to jpsaha@iiserb.ac.in.

⇐⇒ (7k − 5)(4ℓ − 3) = (5k − 3)(6ℓ − 1)

⇐⇒ 28kℓ − 20ℓ − 21k + 15 = 30kℓ − 18ℓ − 5k + 3
⇐⇒ 2kℓ + 2ℓ + 16k − 12 = 0
⇐⇒ kℓ + ℓ + 8k − 6 = 0
⇐⇒ (k + 1)(ℓ + 8) = 14.

This implies that k + 1 is equal to

±1, ±2, ±7, ±14,

i.e. k is equal to
0, 1, 6, 13, −2, −3, −8, −15. (3)
It follows that
(k + 1)(ℓ + 8) = 14
is equivalent to (k, ℓ) being equal to one of the pairs as in Eq. (2). This proves
the Claim.
Note that if a fraction can be written simultaneously in the forms
7k − 5 6ℓ − 1
and
5k − 3 4ℓ − 3
for two integers k, ℓ, then the Claim implies that (k, ℓ) is equal to the pairs as
in Eq. (2), and then k is equal to the integers as in Eq. (3), and consequently,
the fraction 7k−5 6ℓ−1
5k−3 , which is equal to 4ℓ−3 (by the Claim again), is also equal to

5 37 43 19 13 61 30
, 1, , , , , , . (4)
3 27 31 13 9 43 19
Further3 , observe that the preceeding argument also proves that these fractions
can be written simultaneously in the forms as stated above. Indeed, if (k, ℓ) is
one of the pairs as in Eq. (2), and then k is equal to the integers as in Eq. (3),
and consequently, the fraction 7k−5 6ℓ−1
5k−3 , which is equal to 4ℓ−3 (by the Claim), is
also equal to the fractions as in Eq. (4).
We conclude that the fractions as in Eq. (4) are precisely all the fractions
with the required property. ■

3 Note that the argument needs to go on since what we have proved so far does not complete
the solution. The previous step only says that if a fraction can be written simultaneously
in the forms as stated above (and a priori, it is not clear if there is even a single fraction
that can be expressed simultaneously in the stated forms), then the fraction cannot be
anything other than
5 37 43 19 13 61 30
, 1, , , , , , .
3 27 31 13 9 43 19
This does not guarantee if any of these fractions enjoy the stated property.
If this causes any confusion, then it would be a good idea to go through [Che25,
Chapter 1].

Some style files, prepared by Evan Chen, have been adapted here. 5
8 May 2025 https://jpsaha.github.io/MOTP/

Example 1.7 (India RMO 2016e P5).

(i) A 7-tuple (a1 , a2 , a3 , a4 , b1 , b2 , b3 ) of pairwise distinct positive integers
with no common factor is called a shy tuple if

a21 + a22 + a23 + a24 = b21 + b22 + b23

and for all 1 ≤ i < j ≤ 4 and 1 ≤ k ≤ 3, a2i + a2j ̸= b2k . Prove that there
exist infinitely many shy tuples.
(ii) Show that 2016 can be written as a sum of squares of four distinct natural
numbers.

Solution 5. For any integer n ≥ 2,

(3, 4, 3n, 4n, 3(n2 − 1), 6n, 4(n2 + 1))

is a 7-tuple of pairwise distinct positive integers, and these integers have no

common factor. This proves pari (i).
Note that 2016 = 25 · 32 · 7 holds. Observe that 2 · 32 · 7 can be expressed as
42 + 52 + 62 + 72 . This gives

2016 = 25 · 32 · 7 = 162 + 202 + 242 + 282 ,

expressing 2016 as the sum of the squares of four distinct natural numbers. ■

Remark. The identity

(x + y)2 + (x − y)2 = 2(x2 + y 2 )

can also be used to look for shy tuples. Note that

(x − 3y)2 + (x − y)2 + (x + y)2 + (x + 3y)2 = 2(x2 + 9y 2 ) + 2(x2 + y 2 )

= (2x)2 + 20y 2
= (2x)2 + (2y)2 + (4y)2 .

This leads to considering the tuple (x − 3y, x − y, x + y, x − 3y, 2x, 2y, 4y).
To make it a shy tuple, one can take x, y to be positive integers satisfying
x − 3y ≥ 1, and one can impose the condition that x, y are coprime, along with
the additional condition that x, y are of different parity.
Alternatively, one can make use of the following identity.

(a + b + c)2 + (−a + b + c)2 + (a − b + c)2 + (a + b − c)2 = (2a)2 + (2b)2 + (2c)2 .

For more details, we refer to this AoPS thread.

6 The content posted here and at this blog by Evan Chen are quite useful.
References Typos may be reported to jpsaha@iiserb.ac.in.

References
[Che25] Evan Chen. The OTIS Excerpts. Available at https : / / web .
evanchen . cc / excerpts . html. 2025, pp. vi+289 (cited pp. 1, 2,
5)

Some style files, prepared by Evan Chen, have been adapted here. 7