Basic Inequalities and Diophantine Equations

Inequalities and Diophantine Equation Khor Shi-Jie May 15, 2012

Contents 1 Algebraic Inequalities

1.1 1.2 1.3 1.4 1.5

2

General Tips in Solving Simple Inequalities . Quadratic Equations . . . . . . . . . . . . . The AM-GM Inequality . . . . . . . . . . . Completing the Squares . . . . . . . . . . . Problem Set . . . . . . . . . . . . . . . . . .

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2 Diophantine Equations

2.1 2.2 2.3 2.4 2.5 2.6

2 3 3 5 6 7

Considering Residues . . . . . . . . . Factorisation . . . . . . . . . . . . . Discriminant of Quadratic Equation . Bounding and Squeeze Principle . . . Metho d of Infinite Descent . . . . . . Problem Set . . . . . . . . . . . . . .

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7 8 8 9 10 11

Chapter 1 Algebraic Inequalities Looking back after studying MO for several years, the algebraic inequalities involved in junior section 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 which ask for extreme extreme values values or questions questions which ask to prove an algebraic algebraic inequality inequality.. In my opinion, the latter type of problem is easier to solve because at gives you a clearer direction to proceed. 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 tested in junior section. The relevant topics are quadratic equations, AM-GM inequality and algebraic manipulation manipulation skills such as factorisat factorisation ion and completing completing the square. square. Advance Advanced d topics topics in beyond junior section can be helpful too because some problems may be directly derived from these topics (although these problems may still be solved with basic theories).

1.1 1.1

Gene Genera rall Tips Tips in Solvi Solving ng Simp Simple le Ineq Inequa uali liti ties es

1. To prove prove that A

≥ B, consider proving that A − B ≥ 0 or BA ≥ 1.

2. Squares Squares are your friend. Consider Consider completing completing the square. Factorisatio actorisation n can be useful useful too if one side of the inequality is zero. 3. Consider Consider the equality equality case of the inequality inequality.. If the equality equality case occurs o ccurs when all the variables are equal, it is a hint to use AM GM inequality. Note that it is necessary to state the equality case when proving inequalities as the failure to do so will lead to omission of marks.

−

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

≥ ≥

5. Alway Alwayss try and reduce reduce the number number of variabl ariables es in the equati equation. on. This This can be don donee through substitution or clever use of the condition of the problem. Inequalities in one variable can be solved much easily. 2

3

CHAPTER CHAPTER 1. 1. ALGEBRAIC ALGEBRAIC INEQU INEQUALITI ALITIES ES

6. Consider Consider using substitution substitution strategy strategy in solving solving inequalities. inequalities. Some substitutio substitutions ns can simplify the expression in the inequality while some substitutions can take away the condition in the problem. 7. Working orking backw backwards ards is a very very pow powerfu erfull tool in proving proving inequalities. inequalities.

1.2 1.2

Quad Quadra rati tic c Equa Equati tion onss

Quadratic Quadratic equations equations are useful in two two wa ways ys when it comes to extreme values. values. Firstly Firstly,, questions can ask about the minimum minimum or maximum maximum value value of a given quadratic quadratic function. function. This can often often be solved solved through completing completing the square. square. Secondly Secondly,, we can use the discriminan discriminantt of the quadratic quadratic equation equation and the properties properties of its roots to set up an inequalit inequality y. Let us consider the following two examples in evaluating extreme values: Example 1. If the real numbers x, y satisfy the condition 2x2 value of x2 + y 2 + 2x.

− 6x + y

2

= 0, 0 , find the maximal

When there are two variables involved, it is often wise to reduce the number of variables to one. Based on the condition, condition, we can make make the substituti substitution on y 2 = 6x 2x2 into the expression that we want to maximise. Now,

−

x2 + y 2

2

− 2x = −x

2

+ 8x =

−(x − 4) + 16 ≤ 16, which is achieved when x is 4. However, from 2 x − 6x = y ≥ 0 we must have x ≤ 3. This 2

2

means that maximal value of 16 is not achievable and hence the maximum value occurs when x = 3 instead, giving us a value of 15.

Given the number of real roots that a quadratic equation has, we can use the discriminant to constr construct uct an inequa inequalit lity y based based on the coefficien coefficients ts of the quadrati quadraticc equati equation. on. Here’s Here’s an alternative approach to a problem in SMO(J) last year. Exampl Example e 2. (SMO(J) 2011 P1) Suppose a,b,c,d > 0 and x = Prove that xy ac + bd.

≥

√a

2

+ b2 , y =

√c

2

+ d2 .

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

− −

−

4(ac + bd)2

1.3 1.3

2

− 4(a

+ b2 )(c2 + d2 )

≤ 0 ⇔ xy ≥ ac + bd.

The The AMAM-GM GM Ineq Inequa uali litty

Theorem 1 (AM-GM Inequality). For n positive real numbers x1 , x2 ,

· · · x , the arithmetic n

mean of these numbers is larger or equal to the geometric mean of these numbers, that is + xn x1 + x2 + n

···

Equality occurs if and only if x1 = x2 =

≥ √x x · · · x . n

··· = x . n

1

2

n

4


The AM-GM inequality plays a pivotal role in junior section inequalities. It often appears in the following forms: Corollary 1. For two positive real numbers a and b, we have a + b holds if and only if a = b. Corollary 2. For positive real numbers a and b, we have

if and only if a = b.

≥ 2√ab where equality

1

4 ≥ ( a + b) ab

2

where equality holds

The equalit equality y case case of the inequal inequalit ity y is very very important important.. This This is because because if the equalit equality y case cannot be achieved, the inequality will be a strict inequality. To illustrate this concept, consider the expression x2 + 8x1 + x for positive positive x. We cannot conclude that the minimum minimum 3 1 2 value of this inequality is 2 because the equality case x = 8x = x cannot be achiev achieved. ed. We need to concept of calculus to find the minimum point of the above function. 3

3

Here’s an example of AM-GM inequality at work: Exampl Example e 3. (SMO(J) 2011 P1) Suppose a,b,c,d > 0 and x = Prove that xy ac + bd.

√a

2

+ b2 , y =

√c

2

+ d2 .

≥

This same example can be solved using AM that (a2 + b2 )(c2 + d2 ) (ac + bd)2 . We have:

≥

− GM inequa inequalit lity y too.

It suffices suffices to prove prove

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

≥ ≥

where the second last step is derived using the AM-GM inequality. The technique technique of duplication duplication terms to eliminate eliminate variables variables is very very handy. handy. A variable ariable can be duplicated several times in order to eliminate the variables in the denominator, as long as the equality case can be achieved. Consider the following problem: Example 4. Of x > y > 0, find the minimum value of x +

First, we use Corollary 2 to derive that

1 y (x

− y) .

1

4 ≥ . The term in the denominator is y (x − y ) x 2

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

1

1 1 4 ≥ ≥3 x+ x+ 2 2 y (x − y ) x 2

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

5


1.4 1.4

Comp Comple leti ting ng the the Squa Square ress

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 the following problem proposed by Titu Andreescu: Example 5. Let a,b,c be real numbers. Prove that the numbers a

1 cannot be all larger than . 4

2

2

2

2

−b ,b−c ,c−d ,d−a

The solution comes intuitively if you are used to solving inequalities using the method of completing the squares. Let us suppose that it is possible for all for expressions to be larger 1 than at the same time, i.e. 4 a

−b

2

>

1 , 4

b

2

−c

>

1 , 4

c

2

−d

>

1 , 4

d

By adding the four expressions above together, we obtain a+b+c+d

2

− (a

+ b2 + c2 + d2 ) < 1

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

1 2

− a)

2

+(

which is obviously a contradiction.

1 2

− b)

2

+(

1 2

− c)

2

+(

1 2

2

− d)

2

−a

<0

>

1 4

6


1.5

Prob roblem lem Set

1. Find the maximal maximal value of the function 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) (SMO(J) 2010 P35) Suppose the three sides of a triangular triangular field are all integers, integers, and its area equals the perimeter (in numbers). What is the largest possible area of the field? 4. For positive real real numbers a1 , a2 , 2 2 + bn ) (a1 b1 + a2 b2 + b2 +

2 1

2 2

2

2 1

· · · , a , b , b , · · · , b , prove that (a + a +· · ·+ a )(b + ··· ≥ ···+a b ) . 5. For n positive real numbers x , x , · · · x that satisfy x x · · · x = 1, prove that (2 + x )(2 + x ) · · · (2 + x ) ≥ 3 . 1

1

1

n

2

n

n

2

n

n

2

1

n

2

n

n

2

n

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

≤

7. Show that (a2 b + b2 c + c2 a)(ab2 + bc2 + ca2 ) a,b,c. 8. Let x1 , x2 , x1 + x2 +

2 2 2

≥ 9a b c

for all positive real numbers

x2 x2 , xn be positive real numbers. Prove that 1 + 2 + x2 x3 + xn .

··· ···

···

xn2 −1 x2 + + n xn x1

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

2

2

≥ 2

≤ a +b +c .

10. (SMO(J)20 (SMO(J)2009 09 P5) Let a, b be positive real numbers satisfying a + b = 1. Show that if x1 , x2 , , x5 are positive real numbers such that x1 x2 x3 x4 x5 = 1, then (ax1 + b)(ax2 + b) (ax5 + b) 1.

···

···

≥

· · · , x ≤ 1 where n ≥ 1, show that 1 + (nx− 1)x x +···+ ≤ 1. 1 + ( n − 1)x 1

11. (SMO(O (SMO(O)20 )2004 04 P4) If 0 < x1 , x2 , x2 1 + ( n 1)x2

−

n

n

n

12. (SMO(S)19 (SMO(S)1997 97 P1) Let x1 , x2 , x3 , x4 , x5 , x6 be positive real numbers. Show that ( (

x4 5 x6 x1 x3 x5 ) + ( )5 + ( )5 + ( )5 + ( )5 x2 x3 x4 x5 x6

≥ xx

1 2

+

x2 x3 x4 x4 x5 x6 + + + + + . x4 x6 x1 x1 x3 x5

13. Suppose x,y,z are real numbers that satisfy xy + yz + xz = 8z 2 4

≥

+

1

2

−1. Prove that x

x2 5 ) + x1

+ 5y 2 +

14. Given Given that real numbers numbers x, y and z satisfy xy + yz = 10, find the minimum value of 2 2 2 x + 5y + 4z . 15. (Canada (Canada 2012 P1) P1) Let x, y and z be positive real numbers. Show that x2 + xy2 + xyz 2 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 number theory questions in MO competitions. The techniques involved in solving diophantine equations are fairly standard. However, diophantine equation questions can be very creative and studen students ts usuall usually y have have to exhibi exhibitt creati creativit vity y to solve solve these these questi questions ons.. I have have listed listed severa severall techniques that I know of in solving diophantine equations.

2.1 2.1

Cons Consid ider erin ing g Resi Residu dues es

This is one of the most common technique and should be the first technique that students should use to examine a diophantine equation problem. By checking certain common modulos on each term of the equation, one can either arrive at a contradiction to prove that theres no solution, or to find the unique solutions that satisfy the equation. Heres a list of common modulos to take and the possible residues in each scenario: 1. x2 2. x2 3. x2 4. x2 5. x2 6. x2 7. x3 8. x3 9. x4 10. x4

2

≡ 1 (mod 4) when x is odd and x ≡ 0 (mod 4) when x is even. ≡ 0, 1, 4 (mod 8) ≡ 0, 1, 4, 9 (mod 16) ≡ 0, 1 (mod 3) ≡ 0, 1, 4 (mod 5) ≡ 0, 1, 2, 4 (mod 7) ≡ 0, ±1 (mod 7) ≡ 0, ±1 (mod 9) ≡ 0, 1 (mod 5) ≡ 0, 1 (mod 16) 7

8

CHAPTER CHAPTER 2. DIOPHANTIN DIOPHANTINE E EQUATION EQUATIONS S

11. p

≡ ±1 (mod 6) for prime p > 3.

The following problem can be solved using congruence relations: Example 6. Prove that x41 + x42 +

4 14

···+x

= 1983 has no integer solutions.

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

≡

2.2 2.2

Facto actori risa sati tion on

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 terms on the other side of the equations. equations. Then we write write the exponential exponential term ax as am+n where s > t. Upon splitting the exponents, we compare it with the factors on the other side of the equation. Here’s a problem to illustrate this strategy: Example 7. Find all non-negative integer solutions to the equation 3x

−y

3

= 1.

We rewrite the equation as 1 + y 3 = 3x and factorise the left hand side of the equation into (1 + y)(1 y + y 2 ) = 3x . We can set up the following system of equations:

−



y + 1 = 3m y2

where m + n = x and n

≥ m.

−y+1=3

n

It does not seem like we can proceed further. further. If we take a step back, back, we might notice that y cannot be divisible by 3. At this point, we need to introduce introduce another common techniqu techniquee which which complemen complements ts the factorisation factorisation technique: technique: Euclidean Euclidean algorithm. algorithm. Finding Finding the greatest common divisor of the two factors is of tremendous help because the greatest common divisor must be 3m . We have: gcd(y + 1, y 2

+ 2y + 1, y 2 = gcd(3y, y2 y + 1)

− y + 1) = gcd( y

2

−

− y + 1)

which suggests that 3 m 3y . Based on our previous observation that y is not divisible by 3, we can 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

Discri Discrimi minan nantt o off Quad Quadrat ratic ic Equati Equation on

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

9

CHAPTER CHAPTER 2. DIOPHANTIN DIOPHANTINE E EQUATION EQUATIONS S Example 8. (Putnam 1954) Prove that there are no integers x and y such that x2 + 3xy 2y 2 = 122. 122.

−

We have to arrange the equation to appear as a quadratic equation in terms of x. Upon rearrangement we have x2 + 3 xy 2y 2 122 = 0. By taking discriminan discriminant, t, we obtain = 2 17y + 488. Now that you you have been drilled drilled with the idea of taking modulo, it should appear intuitiv intuitivee to take take modulo 17 and check check if it is possibl p ossiblee to be a perfect perfect square. It turns out that 2 17y + 488 12 (mod 17). However, 12 is not a quadratic residue of 17 (can be determined either by listing). Hence the discriminant can never be a perfect square and there will be no integer solutions for ( x, y ).

−

−



≡

2.4 2.4

Boun Boundi ding ng and and Squ Squee eeze ze Pri Princ ncip iple le

Bounding refers to the technique which seeks to find the minimum and maximum bound of a variable ariable and test the feasibility feasibility of the integers integers within the bound individuall individually y. Algebraic Algebraic manipulatio manipulation n skills skills and knowledge knowledge of inequaliti inequalities es are very very useful useful in establishi establishing ng bounds. It is common to make assumptions such as x y if the diophantine equation is symmetric in order to find the bounds for the variables. Here’s an example of this technique:

≥

Example 9. Find all integer solutions to the equation

1 x2

+

1 y2

y

2

which implies that y 2

(1, 1), ( 1, 1).

− −

1 xy

= 1. 1.

≥ y without loss of generality. We then have x1 + y1 + xy1 ≤ ≤ 3. As suc such, h, y = ±1 and hence the solutions are (x, y) =

First, we can assume that x 3

+

2

2

There are many ways in which we can use the squeeze principle, which is a technique whereby whereby the bounds are too tight tight for an integer integer solution to exist. exist. Firstly Firstly,, if we are able to prove that x > a and x < a + 1 for some integer a, then there will definitely be no integer solutions for x. This This simple simple princip principle le can take take many many forms. forms. For example, example, there there will will be no 2 2 2 2 2 integer solutions for m if it is given that m > n and m < (n + 1) . Let us take a look at the following problem from APMO 2011. Example 10. (APMO 2011 P1) Suppose that a,b,c are positive integers. Can the numbers a2 + b + c, a + b2 + c and a + b + c2 be perfect squares at the same time?

Let us suppose that the three numbers numbers can be perfect squares squares at the same time. If that 2 2 2 is the case, since a + b + c > a , we must have a + b + c (a + 1) 2 i.e. b + c 2a + 1. Similarly we have a + b 2c + 1 aand nd a + c 2b + 1. If we were were to add the three inequali inequalities ties together, we obtain 0 3, which is obviously obviously ridiculous. ridiculous. Hence Hence a contradi contradiction ction arises arises and the three numbers cannot be perfect squares at the same time.

≥ ≥

≥

≥

≥


2.5 2.5

10

Meth Method od of of Infin Infinit ite e Desc Descen entt

Method of infinite descent is a method devised by Fermat in solving certain Diophantine equation. We suppose that the equation has non-trivial solutions. Then, we construct smaller solutions solutions from the non-trivial non-trivial solutions and repeat this process ad infinitum. infinitum. Since the set of natural numbers cannot have infinitely small natural numbers, there will be a contradiction which completes the proof. Example 11. Find all integers solutions to the equation x4 + y 4 = z 2

This is a special case of Fermat’ Fermat’ss Last Theorem. Theorem. To prove this, it suffices to show that 4 2 positivee inte integer ger solution solutions. s. We assume assume the contrar contrary y, that that is, a set of x + y = z has no positiv integers (x0 , y0 , z0 ) satisfi satisfies es the equati equation. on. By taking taking modulo modulo 16, we observ observee that that x0 , y0 , z0 are all even numbers and z0 is divisible divisible by 4. We denote x0 = 2x1 , y0 = 2y1 , z0 = 4z1 . This 4 4 suggests that 16x1 + 16 y1 = 16z12 which means that x41 + y14 = z12 . Henc Hence, e, x1 , y1 , z1 is also another another smaller set of solutions. This process can still be repeated repeated ad infinitum. infinitum. Howev However, er, there cannot be infinitely small natural numbers. As such, a contradiction arises and hence then can be no positive integer solutions. 4

11


2.6

Prob roblem lem Set

1. Prove Prove that there are no positive positive integers integers such that a2

− 3b

2

= 8.

2. Prove Prove that there are not positive integers integers such such that x3 + y 3 + 4 = z 3 . 3. Final all positive positive integers integers k, n such such that 1! + 2! +

2

· · · + n! = k .

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

− 8c = 6.

5. (SMO(O) (SMO(O) 1998 P3) Do there there exist integers integers x and y such that 1919 = x3 + y 4 ? Justif Justify y your answer. 6. (USAJMO (USAJMO 2011 P1) Find, with proof, all positive positive integers integers n for which 2n + 12n +2011n is a perfect square. 7. Find all positive positive integers integers such that 2x + 1 = y 2 . 8. (Canada 1972) Prove Prove that the equation x3 +11 3 = y 3 has no solution in positive integers x and y . 9. (SMO(J) (SMO(J) 2008 P5) Determine Determine all primes p such that 5 p + 4 p4 is a perfect square.

·

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

−

11. (Centroam (Centroamerica erican n 2005) Show that the equation equation a2 b2 + b2 c2 + 3b2 no integer solutions.

2

−c −a

2

−

= 2005 has

12. (SMO(J) (SMO(J) 2006 P1) Find all integers integers x, y that satisfy the equation x + y = x2 13. Prove Prove that that x2

− 2xy

2

2

− xy + y .

+ 5z + 3 = 0 has no integer solutions.

14. Find Find all rectang rectangles les with integr integral al sides sides such that the value alue of the area is equal equal to the value of the perimeter. 15. (GDR 1973) Find all integer integer solutions solutions to the equation x(x + 1)(x + 7)(x + 8) = y 2 . 16. (USAMO (USAMO 1976 P3) Determine Determine all integral integral solutions of a2 + b2 + c2 = a2 b2 . 17. (Macedonia (Macedonia 2012) Solve Solve the equation x4 + 2y 4 + 4z 4 + 8t4 = 16xyzt in the set of integer numbers.

Basic Inequalities and Diophantine Equations

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