Mathematical Excalibur, Volume 7, Number 4

Volume 7, Number 4 October 2002 – November 2002

簡 介 費 馬 數

梁 達 榮

Olympiad Corner

The 2002 Canadian Mathematical Olympiad

Problem 1. Let S be a subset of {1, 2, …, 9}, such that the sums formed by adding each unordered pair of distinct numbers from S are all different. For example, the subset {1, 2, 3, 5} has this property, but {1, 2, 3, 4, 5} does not, since the pairs {1, 4} and {2, 3} have the same sum, namely 5. What is the maximum number of elements that S can contain? Problem 2. Call a positive integer n practical if every positive integer less than or equal to n can be written as the sum of distinct divisors of n.

For example, the divisors of 6 are 1, 2, 3, and 6. Since

1 = 1, 2 = 2, 3 = 3, 4 = 1 + 3, 5 = 2 + 3, 6 = 6 we see that 6 is practical.

Prove that the product of two practical numbers is also practical.

(continued on page 4) Editors: 張 百 康(CHEUNG Pak-Hong), Munsang College, HK

高 子 眉 (KO Tsz-Mei) 梁 達 榮 (LEUNG Tat-Wing)

李 健 賢 (LI Kin-Yin), Dept. of Math., HKUST 吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST for general assistance.

On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is

December 15, 2002.

For individual subscription for the next five issues for the 01-02 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:

Dr. Kin-Yin LI Department of Mathematics

The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong

Fax: (852) 2358 1643 Email: [email protected] 考慮形狀如₂m_{+ 1的正整數，如} 果它是質數，則 m 一定是 2 的正次幕。 否則的話，設 m = ₂n_{s，其中 s 是 3 或} 以上的奇數，我們有₂m_{+ 1 = 2}ns 2 + 1 = 2 2 1 2 2 2 _{) 1 ( 2 1)((2 ) (2 )} 2 ( n s+ = n + n s− − n s− ) 1 ± +L ，容易看到₂m_{+ 1分解成兩個} 正 因 子 的 積 。 業 餘 數 學 家 費 馬 (1601-1665)曾經考慮過以下這些“費 馬”整數，設F_n=22n +1, n=0 ,1 ,2 ,...， 費馬看到F0=220+1=3， 2 1 1 2 1= + F = 5，F2=222+1=17， =2 +1= 3 2 3 F 257，F4=224 +1=65537，都是質數， (最後一個是質數，需要花些功夫證 明），他據此而猜想，所有形如22n +1的 正整數都是質數。 不幸的是，大概一百年後，歐拉 (1707 – 1783)發現，F5不是質數，事實 上，直到現在，已知的Fn，n≥5，都 不是質數。對於F5不是質數，有一個 簡單的證明。事實上641=54+24= 1 2 5× 7+ ，因此 641 整除₍₅4_{+2 )2}4 28₌ 4 28 5 ×2 +232。另一方面，由於 641 = 5 1 27+ × ，因此 641 也整除 (5_×27₊₁₎ × = − ×₂7 _{1) 5}2 5 ( 214−1，由此，得到 641 整 除 (52×214−1 )(52×214+1)= × 4 5 228−1。最後 641 整除₅4_×228 32 2 + 和 54×228−1 之 差 ， 即 是 5 32 ₁ 2 + =F 。 這個證明很簡潔，但並不自然，首 先，如何知道一個可能的因子是 641， 其二，641 能夠寫成兩種和式，實有點 幸運。或許可以探究一下，歐拉是怎樣 發現F₅不是質數。我們相信大概的過 程是這樣的，歐拉觀察到，如果p是Fn= 1 22n + 的質因子，則 p 一定是_k⋅2n+1 ₊ 1的形式。用模算術的言語，如果 p 整 除22n +1，則22n ≡−1(modp),取平方， 得出22n+1 ≡1(modp)。另外，用小費 馬 定 理 ，（ 歐 拉 時 已 經 存 在 ）， 知 ) (mod 1 2p−1≡ p 。如果 d 是最小的正整 數，使得2d ≡1(modp)，可以證明（請 自証），d 整除 p – 1，也整除₂n+1_，但 d不整除₂n_{，（因為2}2n _{≡−1(modp))，} 所以_d=2n+1_{，再因 d 整除 p – 1 ，所} 以_p−1=k⋅2n+1_{，或者p=k⋅2}n+1_+1。 （如果用到所謂的二次互反律，還可以 證明，p 實際上是_k⋅2n+2+1_{的形式。）} 例如考慮F4，它的質因子一定是 32k + 1的形式，取 k = 1, 2, …, 等，得 可 能 的 因 子 是 97, 193 ，（ 小 於 65537 ，以 32k + 1 形式出現的質 數）。但 97 和 193 都不整除 65537，所 以 65537 是質數。另外，F5的質因子 一定是 64k + 1 的形式，取 k = 1, 2, …, 等，得可能的因子是 193, 257, 449, 577, 641, …,經 幾 次 嘗 試， 得出225 +1= 4294967297 = 641× 6700417，這樣快就 找出F5的一個質因子，也算幸運，事 實上第二個因子也是質數，不過要證明 就比較麻煩。 但是如果試圖用這樣的方法找尋 其他費馬數的因子，很快就遇上問 題。舉例說，F₆=226 +1是一個二十 位 數 ， 它 的 平 方 根 是 一 個 十 位 數 ) 10 29 . 4 (≈ × 9 ，其中形狀如_k⋅27 _{+ 1 =} 128k + 1的數有三百多萬個，要從中找 尋F6的因子，可不是易事。讀者可以 想像一下，F₅的完全分解歐拉在 1732 年已找到，而在一百年後 Landry 和 Le Lasseur (1880)才 找 到 F6 的 完 全 分 解 ， 再 過 約 一 百 年 ， Morrison 和 Brillhart (1970) 發現F7的完全分解， 因此找尋費馬數的因子分解肯定不是 易事。另一方面由於找尋費馬數不是

Mathematical Excalibur, Vol. 7, No. 4, Oct 02- Nov 02 Page 2 易事，Pepin在1877年找到費馬數是否質 數的一個判斷：N > 3是一個形如22n +1 的費馬數，則N是質數的一個充分必須 條件是 3N2−1 ≡ −1(mod_N)。 考慮到 2 1 − N ₌₂2n−1 ，因此是對3不斷取平方, 然後求對 N 的摸。近代對求費馬數是否 一個質數上，許多都以此為起點。也因 此，曾經有一段長時間，已經知道F₇不 是質數，但它的任一因子都不知道。 再簡述一下近代的結果，現在已知 由F5至F11，都是合數，並且已完全分 解。F₁₂,F₁₃,F₁₅至F₁₉是合數，並且知 道部分因子。但F14,F20,F22等，知道 是合數，但一個因子也不知道。最大的 費 馬 合 數 ， 並 且 找 到 一 個 因 子 的 是 382447 F ，讀者可想像一下，如果以十進 制形式寫下這個數，它是多少個位數。 另外如F33,F34,F35等，究竟是合數或質 數，一點也不知道。有興趣的話，可參 考網頁 http://www.fermatsearch.org/status.htm。 由於費馬數和相關的數有特定的 形式，而且具備很多有趣的性質,因此也 常在競賽中出現。舉例如下： 例一： 給定費馬數F₀,F₁,...,F_n,有以下 的關係F0F1LFn−1+2=Fn。 證明： 事實上Fn=22n +1=22n −1+2 2 ) 1 2 )( 1 2 ( 2 1 22 12− + = 2 1 + 2 1− + = n− n− n− 2 ) 1 2 ( 2 1− ₁+ = − Fn₋ n 。 對於22n−1 −1，可以再分解下去，就可 以得到要求的結果。當然嚴格證明可以 用歸納法。 例二： 給定費馬數 Fm,Fn, m > n，則 n m F F , 是互質的。 證明： 因為Fm=Fm−1LFnLF0+ 2。 設 d 整除 Fm 和Fn, 則 d 也 整除2，所 以d = 1或2。但d≠2，因為F ,m Fn都是 奇數，因此 d = 1，即F ,m Fn互質。 (因此知道，F₀,F₁,F₂,..., 是互質的，即 他們包括無限多數個質因子，引申是有 無限多個質數。) 例三： 有無限多個 n，使得Fn+2不是 質數。 證明： 只要嘗試幾次就可以觀察到F1+ 2 = 7, F3+ 2 = 259, 都是7的倍數。事實 上，對於n = 0, 1, 2, …, ₂2n ≡_{2, 4, 2, 4, …} (mod 7)。因此對於奇數n, Fn+ 2 n 2 2 ≡ + 1 + 2 4≡ + 1 + 2 0≡ (mod 7)， 所以不是質數。 另一個容易看到的事實是： 例四： 對於 n > 1, Fn最尾的數字是 7。 證明: 對於 n > 1, ₂n_{是 4 的倍數，設2}n = 4k, 得 k k n n F =22 +1=24 +1=(24) + 1 ) 5 (mod 2 1 1 + ≡ ≡ k _。因此 n F 最尾的數字 是 2 或 7，它不可以是 2，因為F_n不是 偶數。 例五： 證明存在一個正整數 k，使得對 任何正整數 n，_k⋅2n+1_{都不是質數。} （如果n固定，但容許k在正整數中變 動，由一個重要的定理 (Dirichlet) 知道 在序列中存在無限多個質數。但若果k 固定,而n變動,在序列中究竟有多少個質 數，是否無限多個，一般都不大清楚。 事實上，反可以找到一個k，對於任何正 整數n，k⋅2n+1都不是質數。這原是波 蘭數學家Sierpinski (1882-1969)的一個 結果，後來演變成美國數學奧林匹克 (1982)的一個題目，直到現在，基本是 只有一種證明方法，並且與費馬數有 關。） (續於第四頁)

The 2002 Hong Kong IMO team at the Hong Kong Chek Lap Kok Airport taken on August 1, 2002. From left to right, Chau Suk Ling, Chao Khek Lun, Cheng Kei Tsi, Chiang Kin Nam (Deputy Leader), Yu Hok Pun, Ip Chi Ho, Leung Wai Ying, Li Kin Yin (Leader).

Problem Corner

We welcome readers to submit their solutions to the problems posed below for publication consideration. Solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon. The deadline for submitting solutions is December 15, 2002.

Problem 161.

Around a circle are written all of the positive integers from 1 to N, N ≥ 2, in such a way that any two adjacent integers have at least one common digit in their base 10 representations. Find the smallest N for which this is possible.

Problem 162. A set of positive integers is chosen so that among any 1999 consecutive positive integers, there is a chosen number. Show that there exist two chosen numbers, one of which divides the other.

Problem 163. Let a and n be integers. Let p be a prime number such that p > |a| + 1. Prove that the polynomial f (x) = xn _{+ ax + p cannot be a product of two} nonconstant polynomials with integer coefficients.

Problem 164. Let O be the center of the excircle of triangle ABC opposite A. Let M be the midpoint of AC and let P be the intersection of lines MO and BC. Prove that if

∠

BAC = 2∠ACB, then AB = BP.

Problem 165. For a positive integer n, let S(n) denote the sum of its digits. Prove that there exist distinct positive integers n1, n2, …, n50 such that

. ) ( ) ( ) ( 50 50 2 2 1 1 n S n n S n n S n + = = + = + _L *****************

Solutions

***************** Problem 156. If a, b, c > 0 and , 3 2 2 2_{+b +c =}

a then prove that

. 2 3 1 1 1 1 1 1 ≥ + + + + +ab bc ca

(Source: 1999 Belarussian Math Olympiad)

Solution. SIU Tsz Hang (STFA Leung Kau Kui College, Form 7) and WONG Wing Hong (La Salle College, Form 5). By the AM-GM and AM-HM inequalities, we have ca bc ab + + + + + 1 1 1 1 1 1 2 1 1 2 1 1 2 1 1 2 2 2 2 2 2 _{b b c c a} a + + ₊ + + ₊ + + ≥ . 2 3 3 9 2 2 2_{+ +} = + ≥ c b a

Other commended solvers: CHAN Wai Hong (STFA Leung Kau Kui College, Form 7), CHAN Yat Fei (STFA Leung Kau Kui College, Form 6), CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form 5), CHU Tsz Ying (St. Joseph’s Anglo-Chinese School, Form 7), CHUNG Ho Yin (STFA Leung Kau Kui College, Form 6), KWOK Tik Chun (STFA Leung Kau Kui College, Form 5), LAM Ho Yin (South Tuen Mun Government Secondary School, Form 6), LAM Wai Pui (STFA Leung Kau Kui College, Form 6), LEE Man Fui (STFA Leung Kau Kui College, Form 6),Antonio LEI (Colchester Royal Grammar School, UK, Year 12), LO Chi Fai (STFA Leung Kau Kui College, Form 7), POON Ming Fung (STFA Leung Kau Kui College, Form 5), TAM Choi Nang Julian (SKH Lam Kau Mow Secondary School, teacher), TANG Ming Tak (STFA Leung Kau Kui College, Form 6), TANG Sze Ming (STFA Leung Kau Kui College, Form 5), YAU Chun Biu and YIP Wai Kiu (Jockey Club Ti-I College, Form 5) and Richard YEUNG Wing Fung (STFA Leung Kau Kui College, Form 5).

Problem 157. In base 10, the sum of the digits of a positive integer n is 100 and of 44n is 800. What is the sum of the digits of 3n? (Source: 1999 Russian Math Olympiad)

Solution. CHAN Wai Hong (STFA Leung Kau Kui College, Form 7), CHAN Yat Fei (STFA Leung Kau Kui College, Form 6), Antonio LEI (Colchester Royal Grammar School, UK, Year 12), LO Chi Fai (STFA Leung Kau Kui College, Form 7), POON Ming Fung (STFA Leung Kau Kui College, Form 5), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7), TANG Ming Tak (STFA Leung Kau Kui College, Form 6), and WONG Wing Hong (La Salle College, Form 5).

Let S(x) be the sum of the digits of x in base 10. For digits a and b, if a + b > 9, then S(a + b) = S(a ) + S(b) – 9. Hence, if we have to carry in adding x and y, then S(x + y) < S(x) + S(y). So in general, S(x + y) ≤ S(x) + S(y). By induction, we have S(kx) ≤ kS(x) for every positive integer k. Now

) 40 ( ) 44 ( 800 = S n = S n + n ≤ S(40n)+ S(4n) = 2S(4n) ≤ 8S(n) = 800 .

Hence equality must hold throughout and there can be no carry in computing 4n = n + n + n + n. So there is no carry in 3n = n + n + n and S(3n) = 300. Other commended solvers: CHU Tsz Ying (St. Joseph’s Anglo-Chinese School, Form 7).

Problem 158. Let ABC be an isosceles triangle with AB = AC. Let D be a point on BC such that BD = 2DC and let P be a point on AD such that ∠BAC = ∠BPD. Prove that

∠BAC = 2∠DPC.

(Source: 1999 Turkish Math Olympiad)

Solution. LAM Wai Pui (STFA Leung

Kau Kui College, Form 6), POON Ming Fung (STFA Leung Kau Kui College, Form 5), SIU Tsz Hang (STFA Leung Kau Kui College, Form 7) , WONG Wing Hong (La Salle College, Form 5) and Richard YEUNG Wing Fung (STFA Leung Kau Kui College, Form 5).

Let E be a point on AD extended so that PE=PB. Since ∠CAB=∠EPB and CA/AB=1=EP/PB, triangles CAB and EPB are similar. Then ∠ACB=∠PEB, which implies A, C, E, B are concyclic. So ∠AEC=∠ABC=∠AEB. Therefore, AE bisects ∠CEB.

Let M be the midpoint of BE. By the angle bisector theorem, CE/EB = CD/DB = 1/2. So CE = ½EB = ME. Also, PE = PE and PE bisects ∠CEM. It follows triangles CEP and MEP are congruent. Then ∠BAC = ∠BPE = 2∠MPE = 2∠CPE = 2∠DPC. Other commended solvers: CHAN Yat Fei (STFA Leung Kau Kui College, Form 6), CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form 5) and Antonio LEI (Colchester Royal Grammar School, UK, Year 13).

Mathematical Excalibur, Vol. 7, No. 4, Oct 02- Nov 02 Page 4 Problem 159. Find all triples (x, k, n)

of positive integers such that

. 1 3k _{− = x}n

(Source: 1999 Italian Math Olympiad)

Solution. (Official Solution)

For n = 1, the solutions are (x, k, n) = (3k _{– 1, k, 1), where k is for any positive} integer.

For n > 1, if n is even, then _xn+1≡1

or 2 (mod 3) and hence cannot be 3k _≡ 0 (mod 3). So n must be odd. Now

1 + n x can be factored as ) 1 )( 1 ( ₊ −1 ₋ −2 _{+ +} L n n _x x x .

If 3k _{= x}n _{+ 1, then both of these factors} are powers of 3, say they are 3s_{, 3}t_, respectively. Since x + 1 ≤ xn–1 _{– x}n–2 _{+ +1} L , so s ≤ t . Then 0 ≡ 3t _{≡ (–1)}n–1 _{– (–1)}n–2 _{+ +1} L = n (mod x + 1)

implying n is divisible by x + 1 (and hence also by 3). Let _y=xn/3. _Then

3k _{= y}3 _{+ 1 = (y + 1) (y}2 _{– y + 1).}

So y + 1 is also a power of 3, say it is 3r_. If r = 1, then y = 2 and (x, k, n) = (2, 2, 3) is a solution. Otherwise, r > 1 and

3k _{= y}3 _{+ 1 = 3}3r _{– 3}2r+1 _{+ 3}r+1 is strictly between 33r–1 _{and 3}3r_{, a}

contradiction.

Other commended solvers: LEE Pui Chung (Wah Yan College, Kowloon, Form 7), LEUNG Chi Man (Cheung Sha Wan Catholic Secondary School, Form 6), POON Ming Fung (STFA Leung Kau Kui College, Form 5) and SIU Tsz Hang (STFA Leung Kau Kui College, Form 7). Problem 160. We are given 40 balloons, the air pressure inside each of which is unknown and may differ from balloon to balloon. It is permitted to choose up to k of the balloons and equalize the pressure in them (to the arithmetic mean of their respective pressures.) What is the smallest k for which it is always possible to equalize the pressures in all of the balloons? (Source:1999 Russian Math Olympiad)

Solution. CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form 5) and Antonio LEI (Colchester Royal Grammar School, UK, Year 13).

For k = 5, it is always possible. We equalize balloons 1 to 5, then 6 to 10, and so on (five at time). Now take one balloon from each of these 8 groups. We have eight balloons, say a, b, c, d, e, f, g, h. We can equalize a, b, c, d, then e, f, g, h, followed by a, b, e, f and finally c, d, g, h. This will equalize all 8 balloons. Repeat getting one balloon from each of the 8 groups for 4 more times, then equalize them similarly. This will make all 40 balloons having the same pressure. For k < 5, it is not always possible. If the i-th balloon has initial pressure pi = πi_, then after equalizing operations, their pressures will always have the form c1 p1

+ _{L c}+ 40 p40 for some rational numbers c1, …, c40. The least common multiple of

the denominators of these rational numbers will always be of the form 2r ₃s as k = 1, 2, 3 or 4 implies we can only change the denominators by a factor of 2, 3 or 4 after an operation. So c1, …, c40 can

never all be equal to 1/40.

Olympiad Corner

(continued from page 1) Problem 3. Prove that for all positive real numbers a, b, and c,

c b a ab c ca b bc a3₊ 3₊ 3_{≥ + +}

and determine when equality occurs. Problem 4. Let Γ be a circle with radius r. Let A and B be distinct points on

Γ such that AB < r3 . Let the circle with center B and radius AB meet Γ again at C. Let P be the point inside Γ such that triangle ABP is equilateral. Finally, let the line CP meet Γ again at Q. Prove that PQ = r.

Problem 5. Let N = {0, 1, 2, …}. Determine all functions f:N→N such that ) ( ) ( ) ( ) (y yf x x y f x2 y2 xf + = + +

for all x and y in N.

簡 介 費 馬 數

(續第二頁) 證明：（證明的起點是中國餘式定 理，設m1,m2,...,mr 是互質的正整 數，a₁,a₂,...,a_r是任意整數，則方程 組x≡a₁(modm₁),x≡a₂(modm₂), …, x ≡a_r(modm_r),有解。並且其解 對於模m=m1m2Lm_r唯一。現在考 慮到任意正整數 n，都可以寫成 q₂h 的形式，其中 q 是奇數。如果能夠選 擇 k，使得 k>1,k≡1(mod22h +1), 則 _k⋅2n_+1=k⋅22hq_+1≡(1)(22h₎q h q _{1 ( 1) 1 0(mod2}2 ) 1 )( 1 ( 1≡ − + ≡ − + ≡ + +1)，所以_k2n+1_{不是質數。留意到} 這裡用到 q 是奇數的性質。不過，如 果這樣做的話，h 會因 n 而變，而 k 隨 h 而變，這是不容許的，k 要在起 先之前決定，而不受 n 影響。） 解決的方法是這樣的，我們可以先選 擇 k，使得k>1, k≡1(mod22h +1)， 其中 h = 0, 1, 2, 3, 4。這是可能的， 因為我們知道F₀,F₁,F₂,F₃和

F

4是 不同的質數。這樣的話，可以證明對 於所有n=2hq，其中 h < 5， q 是奇 數 ，k⋅2n+1都 不 是 質 數 。 對 於 q n=2h ,h≥5， 又 可 以 怎樣處 理 呢。留意到所有這樣的數，都可以寫 成n=2hq=25m的形式，其中 m 可 以是奇數，也可以是偶數。另一方 面，我們知道 25 5=2 F + 1 = (641) × (6700417)， 其 中 P = 641 ， Q = 6700417是不同的質數。如果我們選 擇 k，使得 k > 1，和k≡−1(modP)， ) (mod 1 Q k≡ ，則_k2n+1=_k225m+1 m ) 2 )( 1 (− 25 ≡ + 1 ≡(−1)(−1)m+ 1≡ ) (mod 1 ) 1 (− m+1+ P，另一方面_k2n+ m m m k2 1 (1)(2 ) 1 ( 1) 1= 25 + ≡ 25 + ≡ − + 1(mod Q)。如果 m 是偶數，則 1 2n+ k 是 P 的倍數，如果 m 是奇數， 則_k2n+1_{是 Q 的倍數，因此都不是} 質數。歸納言之，選擇 k，使得 ) (mod 1 x k≡ , x = 3, 5, 17, 257, 65537, 6700417, k≡−1(mod641)，則對於 所有形如k2n+1的數，都不是質 數。（最後要留意的是，這方程組的 最小正整數解不可能是 1，因此所有 的k2n+1，都不是質數。）

‘7Solution 1

It can be checked that all the sums of pairs for the set {1,2,3,5,8} are different.

Suppose, for a contradiction, that S is a subset of (1, . . . ,9} containing 6 elements such that all the sums of pairs are different. Now the smallest possible sum for two numbers from S is 1 + 2 = 3 and the largest possible sum is 8 + 9 = 17. That gives 15 possible sums: 3,. . . ,17.

Also there are 6

( 2 > = 15 pairs from S. Thus, each of 3,. . . ,17 is the sum of exactly one pair. The only pair from (1,. . . ,9} that adds to 3 is {1,2} and to 17 is {8,9}. Thus 1,2,8,9 are in S. But then 1+9 = 2 + 8, giving a contradiction. It follows that the maximum number of elements that S can contain is 5.

Solution 2.

It can be checked that all the sums of pairs for the set {1,2,3,5,8} are different.

Suppose, for a contradiction, that S is a subset of (1,. , .9} such that all the sums of pairs are different and that al < aa <‘. . . < ag are the members of S.

Since al + as # a2 + as, it follows that as - a5 # 42 - 41. Similarly as - as # ad - as and

a4 - as # a2 - al. These three differences must be distinct positive integers, so, (as - a5)t(uq - aa)+(az -al) 2 l+ 2 4-3 = 6.

Similarly 43 - as # as - 44, so

(43 - 42) f (us - 44) > 1+ 2 = 3.

Adding the above 2 inequalities yields

ad hence as - al L 9. This is impossible since the numbers in S are between 1 and 9.

Let p and g be practical, For any k 5 pq, we can write k=aq+b with U_<aIp,O<b<q. Since p and q are practical, we can write

a = cr + . . . tcm, b=dl+...+d,,

where the Q’S are distinct divisors of p and the dj’s are distinct divisors of q. NOW k = (cl t . . + cn,)q + (dl + . . . + A)

= qqf.. * +c,,,q+dl+...+d,,.

Each of Gq and dj divides pq. Since dj < q 5 qq for any i,j, the Gq’s and dj’s are all distinct, and we conclude that pq is practical.

Notethat a4 + b4 + c4 = (44 + b4) + (b4fc4) (c4 t u4)

- +

-

.

Applying the arithmetic-geometric

mean inequality to each term2 we see tha? the right ide is greater then or equal to a2b2 + b2c? + c2a2.

We can rewrite this 89

a2(b2 + c2) + b2(c? + u2) + ~?(a~ + b2)

2 2 2

Applying the arithmetic mean-geometric mean inequality again we obtain a4 + b4 + c4 2 a’bc t b2cu t c&b. Dividing both sides by abc (which is positive) the result follows.

Solution 2.

Notice the inequality is homogeneous. That is, if a, b, c are replaced by ku, kb, kc, k > 0 we get the original inequality. Thus we can assume, without loss of generality, that abc = 1. Then

$+;+$ = ,&

( ;+;+z > = a4 + b4 + c4. So we need prove that a4 + b4 + c4 >_ a + b + c.

By the Power Mean Inequality,

a4 + b4 + c4 3 2(q),,

(a + b + c)~ soa4+b4+c4>(a+btc). 27 .

a+b+c

By the arithmetic mean-geometric mean inequality, 3 L~=l,soa+b+c>3. Hence, a4 t b4 + c4 2 (u t b + C) . (4 + 27 b + c)~ _{2(o+b+c);=a+b+c.}

Solution 3.

Rdar than using the Power-Mean inequality to prove 04 + b4 + c4 2 c + b + c in Proof 2,

the Cauchy-Schwartz-Bunjakovsky inequality can be used twice: (a4 + b4 + c4)(12 + 1’ t- 12) 2 (a2 + b2 + .a)2 (c2+b2+c2)(12+12+1*) 1 (a+b+c)2 so a4 + b4 + c4 > (a2 + b2 + c~)~ > (a + b + c)’

3 - 9 _ 81 . Continue as in Proof 2. ..-._. -_

Solution 1.

Let the center of I? be 0, the radius r. Since BP = BC, let 0 = 6BPC = LBCP. Quadrilateral QABC is cyclic, so LBAQ = 180” 10 and hence 6PAQ = 120” - 8.

AhO dAPQ = 180” - LAPB - LBPC = 120” - 8, SO PQ = AQ and L&P = 28 - 60”.

Again because quadrilateral QABC is cyclic, 6ABC = 180” - 6AQC = 240” - 20 ,

Triangles OAB and OCB are congruent, since OA = OB = OC = T a& AB = BC. Thus LABO = 6~~0 = +ABC = 1200 - e.

We have now shown that in triangles AQP and AOB, 6PAQ = LBAO = 6APQ = 6ABO.

Also AP = AB, so AAQP G+ AAOB. Hence QP 4 OB = r.

Solution 2.

Let the center of I’ be 0, the radius r. Since A, P and C he on a circle centered at B, 60” = LABP = 26ACP, so 6ACP = 6ACQ = 30”.

Since Q, A, and C lie on I’, 6QOA = SLQCA = 60”.

So QA = r since if a chord of a circle subtends an angle of 60” at the center, its length is the radius of the circle.

Now BP = BC, so 6BPC = 6BCP = 6ACB + 30”.

Thus 6APQ = 180’ - 6APB - 6BPC = 90’ - LACB.

Since Q, A, B and C lie on I’ and AB = BC, LAQP = 6AQC = LAQB + 6BQC = OLACB.

Finally, 6QAP = 180 - 6AQP - LAPQ = 90 - LACB.

So 6PAQ = LAPQ hence PQ = AQ = r.

l&l&/L

--’

Solution 1. _--

._.

We claim that f is a constant function. Suppose, for a contradiction, that there exist x and y with j(x) < j(y); choose I, y such that f(y) - f(x) > 0 is minimal. Then

f(x) =

e4 +

YfW

<

Xf(Y)

+

Yf(X) < Xf(Y) + Yf(Y)

x+Y x+Y =-l-Y = f(Y)

so f(x) < f(x2 + y2) < f(y) and 0 < f (x2 + y’) - f(x) < f(y) - f(x), contradicting the

choice of x and y. Thus, f is a constant function. Since f(0) is in N, the constant must be from N.

Also, for any c in N, xc +- yc = (x + y)c for all x and y, 80 f(x) = c, c E N are the solutions

to the equation. Solution 2.

We claim f is a constant function. Define g(x) = f(x) - f(0). Then g(6) = 0, g(x) I -f(O) and

xg(y) + yg(x) = (x + y)s(x2 + Y2) for all x, y in N.

Letting y = 0 shows g(x2) = 0 (in particular, g(1) = g(4) = 0), and letting x = y = 1 shows g(2) = 0. Also, if I, y and z in N satisfy x2 + y2 = x2, then

g(Y) =

-$&.

(*I

Letting x = 4 and y = 3, (*) shows that g(3) = 0.

For any even number x = 2n> 4, let y = n2 - 1. Then y > x and z2 + y2 = (n2 + 1)2. For any odd number z = 2nfl > 3, let y = 2(7a+l)n. Then y > 3 and x2+y2 = ((n+1)2+n2)2. Thus for every x > 4 there is y > x such that (*) is satisfied.

Suppose for a contradiction, that there is x > 4 with g(x) > 0. Then we can construct a sequence x = xe < xl < x2 < . . . where g(x;+r) = -z g(x& It follows that /g(x;+r)l > jg(xi)j and the signs ofg(x<) alternate. Since g(x) is alw& an integer, /g(xi+r)l 2 lg(xJ + 1. Thus for some sufficiently large value of i, g(xi) < -f(O), a contradiction.

AS for Proof 1, we now conclude that the functions that satisfy the given functional equation are f(z) = c, c E N.