Inequalities and Diophantine Equation

Khor Shi-Jie

May 15, 2012

Contents

1 Algebraic Inequalities 21.1 General Tips in Solving Simple Inequalities . . . . . . . . . . . . . . . . . . . 21.2 Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 The AM-GM Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Completing the Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.5 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Diophantine Equations 72.1 Considering Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Discriminant of Quadratic Equation . . . . . . . . . . . . . . . . . . . . . . . 82.4 Bounding and Squeeze Principle . . . . . . . . . . . . . . . . . . . . . . . . . 92.5 Method of Infinite Descent . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.6 Problem Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1

Chapter 1

Algebraic Inequalities

Looking back after studying MO for several years, the algebraic inequalities involved in juniorsection is relatively easy to grasp and students who prepare for this topic will be rewarded.There are two likely problems which may come out for algebraic inequalities: questions whichask for extreme values or questions which ask to prove an algebraic inequality. In my opin-ion, the latter type of problem is easier to solve because at gives you a clearer direction toproceed. Still, both kinds of problems can be solved with some practice.

The possible problems which can come out in this topic are limited by the syllabus testedin junior section. The relevant topics are quadratic equations, AM-GM inequality and alge-braic manipulation skills such as factorisation and completing the square. Advanced topicsin beyond junior section can be helpful too because some problems may be directly derivedfrom these topics (although these problems may still be solved with basic theories).

1.1 General Tips in Solving Simple Inequalities

1. To prove that A ≥ B, consider proving that A−B ≥ 0 orA

B≥ 1.

2. Squares are your friend. Consider completing the square. Factorisation can be usefultoo if one side of the inequality is zero.

3. Consider the equality case of the inequality. If the equality case occurs when all thevariables are equal, it is a hint to use AM −GM inequality. Note that it is necessaryto state the equality case when proving inequalities as the failure to do so will lead toomission of marks.

4. For symmetric equations, we can assume that the variables are ordered in some manneri.e. x ≥ y ≥ z. This can help us in bounding the values of the variables.

5. Always try and reduce the number of variables in the equation. This can be donethrough substitution or clever use of the condition of the problem. Inequalities in onevariable can be solved much easily.

2

CHAPTER 1. ALGEBRAIC INEQUALITIES 3

6. Consider using substitution strategy in solving inequalities. Some substitutions cansimplify the expression in the inequality while some substitutions can take away thecondition in the problem.

7. Working backwards is a very powerful tool in proving inequalities.

1.2 Quadratic Equations

Quadratic equations are useful in two ways when it comes to extreme values. Firstly, ques-tions can ask about the minimum or maximum value of a given quadratic function. This canoften be solved through completing the square. Secondly, we can use the discriminant of thequadratic equation and the properties of its roots to set up an inequality. Let us considerthe following two examples in evaluating extreme values:

Example 1. If the real numbers x, y satisfy the condition 2x2−6x+y2 = 0, find the maximalvalue of x2 + y2 + 2x.

When there are two variables involved, it is often wise to reduce the number of variables toone. Based on the condition, we can make the substitution y2 = 6x− 2x2 into the expressionthat we want to maximise. Now,

x2 + y2 − 2x = −x2 + 8x = −(x− 4)2 + 16 ≤ 16,

which is achieved when x is 4. However, from 2x2 − 6x = y2 ≥ 0 we must have x ≤ 3. Thismeans that maximal value of 16 is not achievable and hence the maximum value occurs whenx = 3 instead, giving us a value of 15.

Given the number of real roots that a quadratic equation has, we can use the discriminantto construct an inequality based on the coefficients of the quadratic equation. Here’s analternative approach to a problem in SMO(J) last year.

Example 2. (SMO(J) 2011 P1) Suppose a, b, c, d > 0 and x =√a2 + b2, y =

√c2 + d2.

Prove that xy ≥ ac + bd.

Consider the quadratic equation (a2 + b2)x− 2(ac + bd)x + (c2 + d2) = 0. The quadraticequation can also be rewritten as (ax− c)2 +(bx−d)2 = 0, which suggests that the quadraticequation either has two equal roots or no real roots. The discriminant of the equation is lessthan or equals to 0. Hence,

4(ac + bd)2 − 4(a2 + b2)(c2 + d2) ≤ 0⇔ xy ≥ ac + bd.

1.3 The AM-GM Inequality

Theorem 1 (AM-GM Inequality). For n positive real numbers x1, x2, · · · xn, the arithmeticmean of these numbers is larger or equal to the geometric mean of these numbers, that is

x1 + x2 + · · ·+ xn

n≥ n√x1x2 · · ·xn.

Equality occurs if and only if x1 = x2 = · · · = xn.

CHAPTER 1. ALGEBRAIC INEQUALITIES 4

The AM-GM inequality plays a pivotal role in junior section inequalities. It often appearsin the following forms:

Corollary 1. For two positive real numbers a and b, we have a + b ≥ 2√ab where equality

holds if and only if a = b.

Corollary 2. For positive real numbers a and b, we have1

ab≥ 4

(a + b)2where equality holds

if and only if a = b.

The equality case of the inequality is very important. This is because if the equalitycase cannot be achieved, the inequality will be a strict inequality. To illustrate this concept,consider the expression x2 + 1

8x3 + x for positive x. We cannot conclude that the minimumvalue of this inequality is 3

2because the equality case x2 = 1

8x3 = x cannot be achieved. Weneed to concept of calculus to find the minimum point of the above function.

Here’s an example of AM-GM inequality at work:

Example 3. (SMO(J) 2011 P1) Suppose a, b, c, d > 0 and x =√a2 + b2, y =

√c2 + d2.

Prove that xy ≥ ac + bd.

This same example can be solved using AM − GM inequality too. It suffices to provethat (a2 + b2)(c2 + d2) ≥ (ac + bd)2. We have:

(a2 + b2)(c2 + d2) = (ac)2 + (bd)2 + a2d2 + b2c2

≥ (ac)2 + (bd)2 + 2(ac)(bd)

≥ (ac + bd)2

where the second last step is derived using the AM-GM inequality.

The technique of duplication terms to eliminate variables is very handy. A variable canbe duplicated several times in order to eliminate the variables in the denominator, as long asthe equality case can be achieved. Consider the following problem:

Example 4. Of x > y > 0, find the minimum value of x +1

y(x− y).

First, we use Corollary 2 to derive that1

y(x− y)≥ 4

x2. The term in the denominator is

of degree 2. To eliminate this term, we need to duplicate the term x in the expression thatwe want to minimise. We have:

x +1

y(x− y)≥ 1

2x +

1

2x +

4

x2≥ 3

by AM-GM inequality. We note that equality is achieved when x = 2 and y = 1. Hence theminimum value is 3.

CHAPTER 1. ALGEBRAIC INEQUALITIES 5

1.4 Completing the Squares

The technique of completing the squares is a fundamental technique in proving inequalities.This is based on the property that no square numbers are positive. Let us take a look at thefollowing problem proposed by Titu Andreescu:

Example 5. Let a, b, c be real numbers. Prove that the numbers a− b2, b− c2, c− d2, d− a2

cannot be all larger than1

4.

The solution comes intuitively if you are used to solving inequalities using the method ofcompleting the squares. Let us suppose that it is possible for all for expressions to be larger

than1

4at the same time, i.e.

a− b2 >1

4, b− c2 >

1

4, c− d2 >

1

4, d− a2 >

1

4

By adding the four expressions above together, we obtain

a + b + c + d− (a2 + b2 + c2 + d2) < 1

Moving all terms to the right and completing the squares, we have

(1

2− a)2 + (

1

2− b)2 + (

1

2− c)2 + (

1

2− d)2 < 0

which is obviously a contradiction.

CHAPTER 1. ALGEBRAIC INEQUALITIES 6

1.5 Problem Set

1. Find the maximal value of the function y = x|x| − 2x + 1 when |x + 1| ≤ 6.

2. (SMO(O) 2009 P8) It is given that√a −√b = 20, where a and b are real numbers.

Find the maximum possible value of a− 5b.

3. (SMO(J) 2010 P35) Suppose the three sides of a triangular field are all integers, and itsarea equals the perimeter (in numbers). What is the largest possible area of the field?

4. For positive real numbers a1, a2, · · · , an, b1, b2, · · · , bn, prove that (a21+a22+· · ·+a2n)(b21+b22 + · · ·+ b2n) ≥ (a1b1 + a2b2 + · · ·+ anbn)2.

5. For n positive real numbers x1, x2, · · ·xn that satisfy x1x2 · · ·xn = 1, prove that (2 +x1)(2 + x2) · · · (2 + xn) ≥ 3n.

6. Given that (1 + a)(1 + b)(1 + c) = 8 where a, b, c are positive real numbers, prove thatabc ≤ 1.

7. Show that (a2b + b2c + c2a)(ab2 + bc2 + ca2) ≥ 9a2b2c2 for all positive real numbersa, b, c.

8. Let x1, x2, · · · , xn be positive real numbers. Prove thatx21

x2

+x22

x3

+ · · · +x2n−1

xn

+x2n

x1

≥x1 + x2 + · · ·+ xn.

9. Let a, b, c be positive real numbers that satisfy abc = 1. Prove that a+b+c ≤ a2+b2+c2.

10. (SMO(J)2009 P5) Let a, b be positive real numbers satisfying a + b = 1. Show that ifx1, x2, · · · , x5 are positive real numbers such that x1x2x3x4x5 = 1, then (ax1 + b)(ax2 +b) · · · (ax5 + b) ≥ 1.

11. (SMO(O)2004 P4) If 0 < x1, x2, · · · , xn ≤ 1 where n ≥ 1, show thatx1

1 + (n− 1)x1

+

x2

1 + (n− 1)x2

+ · · ·+ xn

1 + (n− 1)xn

≤ 1.

12. (SMO(S)1997 P1) Let x1, x2, x3, x4, x5, x6 be positive real numbers. Show that (x2

x1

)5 +

(x4

x2

)5 + (x6

x3

)5 + (x1

x4

)5 + (x3

x5

)5 + (x5

x6

)5 ≥ x1

x2

+x2

x4

+x3

x6

+x4

x1

+x4

x1

+x5

x3

+x6

x5

.

13. Suppose x, y, z are real numbers that satisfy xy + yz +xz = −1. Prove that x2 + 5y2 +8z2 ≥ 4

14. Given that real numbers x, y and z satisfy xy + yz = 10, find the minimum value ofx2 + 5y2 + 4z2.

15. (Canada 2012 P1) Let x, y and z be positive real numbers. Show that x2+xy2+xyz2 ≥4xyz − 4.

Chapter 2

Diophantine Equations

A diophantine equation is a equation or a system of equations with multiple variables sub-jected to the condition that the variables are integers. It contributes to a large bulk of numbertheory questions in MO competitions. The techniques involved in solving diophantine equa-tions are fairly standard. However, diophantine equation questions can be very creative andstudents usually have to exhibit creativity to solve these questions. I have listed severaltechniques that I know of in solving diophantine equations.

2.1 Considering Residues

This is one of the most common technique and should be the first technique that studentsshould use to examine a diophantine equation problem. By checking certain common moduloson each term of the equation, one can either arrive at a contradiction to prove that theresno solution, or to find the unique solutions that satisfy the equation. Heres a list of commonmodulos to take and the possible residues in each scenario:

1. x2 ≡ 1 (mod 4) when x is odd and x2 ≡ 0 (mod 4) when x is even.

2. x2 ≡ 0, 1, 4 (mod 8)

3. x2 ≡ 0, 1, 4, 9 (mod 16)

4. x2 ≡ 0, 1 (mod 3)

5. x2 ≡ 0, 1, 4 (mod 5)

6. x2 ≡ 0, 1, 2, 4 (mod 7)

7. x3 ≡ 0,±1 (mod 7)

8. x3 ≡ 0,±1 (mod 9)

9. x4 ≡ 0, 1 (mod 5)

10. x4 ≡ 0, 1 (mod 16)

7

CHAPTER 2. DIOPHANTINE EQUATIONS 8

11. p ≡ ±1 (mod 6) for prime p > 3.

The following problem can be solved using congruence relations:

Example 6. Prove that x41 + x4

2 + · · ·+ x414 = 1983 has no integer solutions.

It is common to assess the residue mod 16 if the equation contains terms of degree 4. Bytaking mod 16, we realise that 1983 gives a residue of 15 when divided by 16, but the factthat x4 ≡ 0, 1 (mod 16) shows that it is impossible for the left hand side of the equation tobe equal to 15 mod 16.

2.2 Factorisation

This is another common technique especially when exponential terms such as ax are given.We place the exponential term on one side of the equation and factorise the other termson the other side of the equations. Then we write the exponential term ax as am+n wheres > t. Upon splitting the exponents, we compare it with the factors on the other side of theequation. Here’s a problem to illustrate this strategy:

Example 7. Find all non-negative integer solutions to the equation 3x − y3 = 1.

We rewrite the equation as 1 + y3 = 3x and factorise the left hand side of the equationinto (1 + y)(1− y + y2) = 3x. We can set up the following system of equations:{

y + 1 = 3m

y2 − y + 1 = 3n

where m + n = x and n ≥ m.

It does not seem like we can proceed further. If we take a step back, we might notice thaty cannot be divisible by 3. At this point, we need to introduce another common techniquewhich complements the factorisation technique: Euclidean algorithm. Finding the greatestcommon divisor of the two factors is of tremendous help because the greatest common divisormust be 3m. We have:

gcd(y + 1, y2 − y + 1) = gcd(y2 + 2y + 1, y2 − y + 1)

= gcd(3y, y2 − y + 1)

which suggests that 3m|3y. Based on our previous observation that y is not divisible by 3, wecan conclude that either m = 0 or m = 1. When m = 0, we obtain the solution (x, y) = (0, 0)and when m = 1, we obtain the solution (x, y) = (2, 2). These are the only 2 solutions.

2.3 Discriminant of Quadratic Equation

If the diophantine equation given is a quadratic equation with 2 variables, the discriminantcan come in handy to determine the bounds of the variables. If the variables are stated to bepositive integer, it suggests that the discriminant must be a perfect square (or in rare cases,the square of a rational number). Here’s an old problem from Putnam, an undergraduatecompetition for students in American universities:

CHAPTER 2. DIOPHANTINE EQUATIONS 9

Example 8. (Putnam 1954) Prove that there are no integers x and y such that x2 + 3xy −2y2 = 122.

We have to arrange the equation to appear as a quadratic equation in terms of x. Uponrearrangement we have x2 + 3xy − 2y2 − 122 = 0. By taking discriminant, we obtain 4 =17y2 + 488. Now that you have been drilled with the idea of taking modulo, it should appearintuitive to take modulo 17 and check if it is possible to be a perfect square. It turns out that17y2 + 488 ≡ 12 (mod 17). However, 12 is not a quadratic residue of 17 (can be determinedeither by listing). Hence the discriminant can never be a perfect square and there will be nointeger solutions for (x, y).

2.4 Bounding and Squeeze Principle

Bounding refers to the technique which seeks to find the minimum and maximum bound ofa variable and test the feasibility of the integers within the bound individually. Algebraicmanipulation skills and knowledge of inequalities are very useful in establishing bounds. Itis common to make assumptions such as x ≥ y if the diophantine equation is symmetric inorder to find the bounds for the variables. Here’s an example of this technique:

Example 9. Find all integer solutions to the equation1

x2+

1

y2+

1

xy= 1.

First, we can assume that x ≥ y without loss of generality. We then have1

x2+

1

y2+

1

xy≤

3

y2which implies that y2 ≤ 3. As such, y = ±1 and hence the solutions are (x, y) =

(1,−1), (−1, 1).

There are many ways in which we can use the squeeze principle, which is a techniquewhereby the bounds are too tight for an integer solution to exist. Firstly, if we are able toprove that x > a and x < a + 1 for some integer a, then there will definitely be no integersolutions for x. This simple principle can take many forms. For example, there will be nointeger solutions for m2 if it is given that m2 > n2 and m2 < (n + 1)2. Let us take a look atthe following problem from APMO 2011.

Example 10. (APMO 2011 P1) Suppose that a, b, c are positive integers. Can the numbersa2 + b + c, a + b2 + c and a + b + c2 be perfect squares at the same time?

Let us suppose that the three numbers can be perfect squares at the same time. If thatis the case, since a2 + b + c > a2, we must have a2 + b + c ≥ (a + 1)2 i.e. b + c ≥ 2a + 1.Similarly we have a + b ≥ 2c + 1 and a + c ≥ 2b + 1. If we were to add the three inequalitiestogether, we obtain 0 ≥ 3, which is obviously ridiculous. Hence a contradiction arises andthe three numbers cannot be perfect squares at the same time.

CHAPTER 2. DIOPHANTINE EQUATIONS 10

2.5 Method of Infinite Descent

Method of infinite descent is a method devised by Fermat in solving certain Diophantineequation. We suppose that the equation has non-trivial solutions. Then, we construct smallersolutions from the non-trivial solutions and repeat this process ad infinitum. Since the set ofnatural numbers cannot have infinitely small natural numbers, there will be a contradictionwhich completes the proof.

Example 11. Find all integers solutions to the equation x4 + y4 = z2

This is a special case of Fermat’s Last Theorem. To prove this, it suffices to show thatx4 + y4 = z2 has no positive integer solutions. We assume the contrary, that is, a set ofintegers (x0, y0, z0) satisfies the equation. By taking modulo 16, we observe that x0, y0, z0are all even numbers and z0 is divisible by 4. We denote x0 = 2x1, y0 = 2y1, z0 = 4z1. Thissuggests that 16x4

1 + 16y41 = 16z21 which means that x41 + y41 = z21 . Hence, x1, y1, z1 is also

another smaller set of solutions. This process can still be repeated ad infinitum. However,there cannot be infinitely small natural numbers. As such, a contradiction arises and hencethen can be no positive integer solutions.

CHAPTER 2. DIOPHANTINE EQUATIONS 11

2.6 Problem Set

1. Prove that there are no positive integers such that a2 − 3b2 = 8.

2. Prove that there are not positive integers such that x3 + y3 + 4 = z3.

3. Final all positive integers k, n such that 1! + 2! + · · ·+ n! = k2.

4. (Canada 1969) Show that there are no integers a, b, c for which a2 + b2 − 8c = 6.

5. (SMO(O) 1998 P3) Do there exist integers x and y such that 1919 = x3 + y4? Justifyyour answer.

6. (USAJMO 2011 P1) Find, with proof, all positive integers n for which 2n +12n +2011n

is a perfect square.

7. Find all positive integers such that 2x + 1 = y2.

8. (Canada 1972) Prove that the equation x3+113 = y3 has no solution in positive integersx and y.

9. (SMO(J) 2008 P5) Determine all primes p such that 5p + 4 · p4 is a perfect square.

10. (SMO(S) 2011 P32) It is given that p is a prime number such that x3 +y3−3xy = p−1for some positive integers x and y. Determine the largest possible value of p.

11. (Centroamerican 2005) Show that the equation a2b2 + b2c2 + 3b2 − c2 − a2 = 2005 hasno integer solutions.

12. (SMO(J) 2006 P1) Find all integers x, y that satisfy the equation x+ y = x2−xy + y2.

13. Prove that x2 − 2xy2 + 5z + 3 = 0 has no integer solutions.

14. Find all rectangles with integral sides such that the value of the area is equal to thevalue of the perimeter.

15. (GDR 1973) Find all integer solutions to the equation x(x + 1)(x + 7)(x + 8) = y2.

16. (USAMO 1976 P3) Determine all integral solutions of a2 + b2 + c2 = a2b2.

17. (Macedonia 2012) Solve the equation x4 + 2y4 + 4z4 + 8t4 = 16xyzt in the set of integernumbers.