數學女王頭上的皇冠

熱身題 先來熱身一下,證明一個簡單的題目: 證明素數的平方根是無理數. 請各位數論迷試試.

請先問問張兄 甚麼是素數？可否解釋一下數論其實是甚麼，有何重要性？謝謝．

難題 請各位數學迷試證如下題目: a, b是正整數, 且a^2+b^2能被1+ab整除, 證明: (a^2+b^2) /(1+ab) 是一個完全平方數. 或者說證明: (a^2 + b^2) ≡ 0 mod(1+ab)=> [(a^2 + b^2) /(1+ab)] = K^2 其中 a,b,K εN 這道數論題較難,首先證明出來的人,我送他一本中文書(可以一起到三聯書店或商務印書館選購)兼請他吃一餐飯. 由於我最近有要務在身,將暫別這個網站,或只是偶爾露一下臉,不會參與討論,對錯誤的證明不做回應. 希望各位能互相研究.

所謂「要務在身」,是寫你提過要出版的書？

送一張"無間道"戲飛好冇?

Answer I think JJ's answer is mostly correct, but incomplete. I can't type those fancy powers, so I will use the Pascal- or C- type symbols as follows. First, I give some counter-examples for JJ. I didn't do it by pencil and paper, I wrote a simple program. The following pairs are all solutions of (a,b) < 100, with symmetric ones removed: a=1 b=1 a=2 b=8 a=3 b=27 a=4 b=64 a=8 b=30 JJ showed successfully that when a=b, the only possiblity is a=b=1: 2a^2=n(1+a^2) => (2-n)(a^2)=n => n=1 (since otherwise (2-n) = 0 or (2-n) < 0) but (a^2) is positive, if (2-n) is negative, n=(2-n)(a^2) will be negative. So n=1,2a^2=n(1+a^2) => a^2=1 => a=1. He missed the point that the statement can still be true even if a, b are not the same. When a!=b (@) if a1 and a2 are roots of a in a^2 - nba + b^2 - n = 0 (*) (NBA!) then a2 = nb - a1. Since both a1 and a2 fulfil (*), that is, if a=alpha fulfils (*), so is (nb-alpha). And the same is true for b too, because a and b are symmetric, i.e. if b=beta fulfils (*), so is (na-beta). Therefore, we have a recurrence relation of pairs of integers fulfilling (*): _0 = a _1 = b, _k = n (_k-1) - (_k-2) such that p(k): "_k^2 + (_k+1)^2 = n (1 + _k _k+1)" We know _0 and _1 are both not zero (from (@)) And if _k-1, _k are BOTH NOT zero, _k, _k+1 are NOT BOTH zero. Since the sequence is strictly monotonic decreasing, if _k+1 is zero, _k+2 will be negative. Hence we've proved by MI that not both _k and _k+1 in any p(k) are zero. Since the sequence is strictly monotonic decreasing, there must exist a k' such that in p(k') _k' > 0 but _k'+1 <= 0. (For a finite integer, everytime it decreases by a certain amount, it will pass zero at some large k. ) we prove _k'+1 won't be < 0 by indirect proof: p(k'): _k'^2 + (_k'+1)^2 = n (1 + _k' _k'+1) n = (_k'^2 + (_k'+1)^2) / (1 + _k' _k'+1) And assume _k'+1 < 0, if _k'+1 = 1, n is undefined otherwise n would be negative(numerator always positive, denominator negative). This is impossible. So _k'+1 = 0. if _k'+1 = 0, _k'^2 + (_k'+1)^2 = n (1 + _k' _k'+1) => _k'^2 = n n is a perfect square! (Q.E.D.) Yes, this is a very hard question.

Why not the warm-up one too? By indirect proof: Assuming that the statement were not true, sqrt(p) = a/b a, b: coprime integers (if not, factor the common factors out) p=a^2/b^2 p(b^2)=a^2 p is a factor of a^2 by unique factorization theorem, p must be a factor of a. Write a=pk (sure, why not?) p(b^2)=(pk)^2 b^2=p k^2 Similarly, p is a factor of b. Well, p is both factor of a and b, a and b are not coprime. This is a contradiction, as JPY would also agree, the statement is true. :)

Questions? To S.C. You have not proved that all [_(K),_(K+1)] dervied by _0 = a _1 = b, _K = n* _(K-1) - _(K-2) ---(1) satisfied p(K): "_(K)^2 + _(K+1)^2 = n (1 + _(K) _(K+1))" ---(2). If p(K) is a part of definitation, how do you know all _(K), stisfied (1) and (2), exist if k>1 and _(K) are positive integers?

哈哈, 變左數學板 :P 有趣的是另一個數字題目開始變政治板 :P 好想學李天命先生, 對住S.C. 講 bravo! 嘻嘻 不過S.C. 你應睇多次我那個proof. 其實我有考慮a!=b 的情形 ( WLOG, 即 without loss of generality, 我考慮的是 a > b ) 另外, 你的反例是有效的. 主要是因為原題目打少左一個 condition (a,b)=1 即要求a,b 互質). 通常做數論的題目, 這是基本. 因為如果a,b有公因子r, 則r² 乘原題目的解會也是明顯/直觀解 (當然我們已知原題目的解是1, 所以(a,b)!=1時, r² 會是新題目的解) 我還以為這裡的既有人自稱數學迷/狂/人, 應該會認識費馬發明的"無窮下降法"吧? 到頭來要勞煩S.C.老哥你幫我補充如何及什麼是"無窮下降法" 再一次唔該你~~ :)

S.C. 老兄~ 唔記講... 下次你想打平方或其他數學符號, 可以安裝相關的字体and/or 輸入法 簡單的一個半個符號也可以從Character Map 抄出 (Character Map 通常在 progarm file>accessories>system tools 可找到) 以前也可以按alt-num### , 不過似乎某些system /console 未必出到.

問JJ 或其他網友 JJ 如唔, 還記得我在本網區的什麼地方表露過很欣賞你嗎？我現在有個問題：怎樣用歸謬法證明矛盾命題不成立？請你或其他網友為我解惑, 謝謝！

JJ 是指我吧? 欣賞不欣賞不是一個我會回應你的理由吧? (我反而較想回應一些要和我鬥咀的高手高人.....) 反正你就算不欣賞我, 但問題問得好我也會有興趣答 你的問題是"怎樣用歸謬法證明矛盾命題不成立？" 嗎? 你可以講講對此有何疑惑呢??

我不是問得很清楚了嗎：怎樣用歸謬法證明矛盾命題不成立？

玲兒~ 不是耍你.... 我知你的"問題"是什麼, 我是問你對此"有何疑惑" ?? 你的意思是不應成立? 或者你認為要睇情形? 有時成立, 有時不成立? 我同你未至於心靈相通吧? ( :P ) 我點知你的疑惑是什麼???

唉！你能否用歸謬法證明矛盾命題不成立？

這𥚃有無好心人呢？ 無窮下降法是什麼東西啊！康慈忘記掉了！可以用中文說明嗎？ 康慈想當年[1988年左右]去北京大學玩時！二三下解了一個別人解不了的數學公式！康慈又忘了那是什麼公式！有無好心把它人登出來啊！ 康慈出生時腦缺氧，加上又有多次重度失憶的記錄！康慈記得自已有四個或以上的哲學思想：烏龜哲學、白兔哲學、開心哲學、康慈哲學或帝王學！康慈好在記得自已最重要的哲學：烏龜哲學！康慈很少和別人說烏龜哲學的內容！[好在自己記得]有沒好心人、十三點或[李天命]先生等有記性的好心人把康慈的白兔哲學、開心哲學、康慈哲學或帝王學登上網呢？

JoeJoneS 不知所云(不是說你，是說旁邊那位)。你正在要證明給我看嗎？還在等你呢！

(1) p&~p [Hyp] (2) ~(p&~p) [RAA, Hyp discharged]

有趣有趣~~ 第一次不用一開始就用丈八金剛 康慈, 是的, 是第6題 :P 你也玩過IMO嗎? 那一年? (可惜始終要送個丈八金剛給你... 你之後講的我不明白.... 60_0 玲兒, 我在新聞組中研究緊問題, 遲一點再和你相會.... 還有這位有趣的康慈 :-目

噢... 對不起... 睇漏了"80年代李學徒"前輩... 我也想會一會你 :)

To Testing Testing: At your service. //To S.C. You have not proved that all [_(K),_(K+1)] dervied by _0 = a _1 = b, _K = n* _(K-1) - _(K-2) ---(1) satisfied p(K): "_(K)^2 + _(K+1)^2 = n (1 + _(K) _(K+1))" ---(2). // First, you understand that a^2 + b^2 = n (1 + ab) (A) => a'^2 + b'^2 = n (1 + a'b'), (B) where a' = b, b' =nb - a. Right? Now a,b are arbitrary. No matter what they are, if they fulfil (A), a', b' must fulfil (B). So we have a recurrence relation. To make it simpler, define proposition p'(x,y) = 'x^2 + y^2 = n (1 + xy)' with all given requirements I have (well JJ has) showed that p'(a, b) => p'(b, nb - a), for all a and b. (C) If (a,b) fulfils the antecedent, the consequence will be also true. p(k) in (2) just means p'(_k, _l+1), right? I tackle the recurrence by MI, of course. p'(0, 1): given. (Strictly speaking, the convergence of the sequence { _k } should take an assumption as JJ suggested a > b. This is okay because a, b are arbitrary and symmetric. ) Assume p(_k-1, _k) is true. (k >= 1) By (C), p(_k, n(_k) - _k-1) (just subst _k-1, _k into (C)) By (1), p(_k, _k+1) (simply applying (1) for _k+1) So by MI (without those formality cuties), p'(_k, _l+1), and its equivalence p(k) are true. Can I claim my prize? :)

To JoeJones Well, of course I know you tried to tackle the a>b case. So I said what I said. :) Heehee, I don't type symbols so often so I will pass, but thanks for your info. If I type, I will use latex or just word. I guess people are not familiar with chinese terms? :) Thank you.

To Great Mathematician, s.c. a and b cannot be are arbitrary since you need to fix the value of n. {_0=a, _1=b} of course fulfills (1) and (2) because they are definition. However what is the case when k>1? Let’s use your findings, counter example of JoeJones’s proof, which show JoeJone’s proof incorrect. Let’s say _0=a=8 _1=b=2 => n=2 according to (1), _2=2*2-8=-4 _3=2*(-4)-2=-10 . . . Now let’s check if {_1,_2} and {_2,_3} fulfill (2), For case {_1=2, _2=-4} LHS=_1^2+_2^2 =(2)^2+(-4)^2 =20 RHS=n*(1+_1*_2) =2*(1+2*(-4)) =-14 =/=LHS similarly, for case {_2=-4,_3=-10} LHS=116=/=RHS=82 This simple calculation shows when k>1, all pairs {_K,_(K+1)} derived by (1) cannot satisfy (2). That means your definition is not self consistent. Therefore, your proof is definitely WRONG!

To Great Arithmetician TESTING When a = 8, b = 2, n = (64 + 4) / (1 + 16) = 4. Thank you.

To Great Mathematician, s.c. Again Thank you for pointing out my mistake!!! This is an unforgivable fault, so I decided to study Arithmetic again as well as attending primary school classes.

太好野 la ！ TESTING ： // This is an unforgivable fault, so I decided to study Arithmetic again as well as attending primary school classes. // 你咁講真係好o勒！我以後就乜數都唔駛睇、唔駛讀、唔駛計咁話 lah ！多謝哂你 ah ，阿 TESTING ！

多謝80年代李學徒 多謝,好多謝！

Do we have any pure math major here?

Are you the founder of Robinson's arithmetic and non-standard analysis？

潛存讀者 今天才有空上網,遲了回答,抱歉. 所謂"要務在身",是我將最近的研究成果總結一下,打算將之上網與大家討論.我將新開一個話題----"分析邏輯". 謝謝!

To S.C. 我剛將你的證明印出來(看電腦眼睛較辛苦),讓我仔細研究一下.

向S.C.喝采！！！ 初步看一下，你的證明應該是正確的，先向你致敬． 由於有些數學符號無法直接打在留言區裏，你就使用電腦上的數學語言，但我不熟悉這些語言，有些符號我要靠猜測． 你能否用Latex(.ps)或word(.doc)將完整詳盡的證明用標準的數學語言打成文件擺上網，讓我們可以下載？ 你的證明十分優美．愛好數學的人不但要求有一個證明，而且要求證明優美簡潔．我相信許多愛好數論的朋友都有興趣下載並珍藏你優美的證明． 約個時間領取你的獎品，謝謝！

JoeJones 佩服你的數學能力，惜邏輯性稍欠嚴謹，而且粗枝大葉．如果沒有這些缺點，你當年就會是香港代表隊的正選． 無論如何，你對這個證明也有貢獻．S.C.領全份獎品，你領半份----請你吃一餐，如何？不吃白不吃．

n/t http://www.geocities.com/xxh33/proof.gif I didn't have time to check the details. Let me know if there are any typos or mistakes. :)

入學啦! 十三點, 我喺春田花花小學讀一年班,喺1A班學算術,你喺唔喺我ge同學仔吓?下學期見啦!記住到時借功課我抄喎! Bye Bye

寫緊書 十三點去春田花花幼兒園度面試 nursery 班，點知道個度 d人話我個人頭豬腦唔夠大，未夠斤兩，仲即刻叫我走添 ah！真真令十三點好開心呀！ 唔係o羊？唔駛返學喎！你估做阿麥 X 同麥 Y 個同學仔好咩？我都話左最驚 d 數架 lah！你係唔係嚮度笑人地 jeh？你阿 TESTING 都唔多理人地個心 ge！ 我同阿 S. C. 講過話整理緊自己寫 ge 野，遲 d 會出本《十三點 D思考藝術》，一於做思想家咁話架！ 思想家唔駛考牌個播，又唔駛讀書， 你唔知 ge咩？

也可以再用其他方法證明 思方迷（也許還有Robinson），看樣子你（們）是讀純數學的．看過S.C.的證明了嗎？太美妙了．你興趣也可以以不同的方法再證明一次．其他人也是．

更正 "你興趣"應是"你有興趣".

給數學家S.C. 謝謝您將完整的證明寫出來．我將它印出來，對它進行最後的驗證，從中獲得極大的樂趣． 近來專注於分析邏輯的研究和建構，少看數學書，好久沒有欣賞到這樣美妙的證明．再次謝謝您． 有幸請您吃飯嗎？如果您是在大學教書，我可以到學校去找您，這樣就不會浪費您太多時間，也可以順便在大學書店選購您感興趣的書． 我的電郵：nought@ctimail.com

上路 . 忘情 !你出書啊!可唔可以送本俾我, 我都想學下十三點ge思考方法學。 其實我都想出本書來幫下世間ge情痴,書名我諗左係<<上路 . 忘情>> 或者 係<<忘情 . 上路>>,你話邊個名好? 為左寫呢本書,聽日要走啦!去感受一下上路及忘情。 有緣再見!

對待錯誤的五種方式 當自己錯誤的觀點被人指出後，應該怎麼辦？ 下下策：破口大罵，咀咒對方下地獄．如揣摩． 下策：詭辯，如梁燕城博士． 中策：做無頭龜，如張海澎． 上策：承認錯誤，如愛因斯坦． 上上策：自嘲，如Testing.

SC: Nice proof! Except that in line 2 of case 2 you might want to change the =>'s to <=>'s. 張海澎: I can't think of a different proof right now, but it's something interesting to think about.

To ： TESTING 乜話 ~ ？拿，你自己打自己o既咀架咋，唔好一陣間又入我數 ah ！ 又話自己要忘情，咁做o羊野重要問我播？我講黎又做o羊播？

上路 再見!

To Robinson (P<=>q)=>(p=>q) "=>" is enough for the purpose of this proof.

張海澎 I think he needs the fact that a is a root for x^2-(nb)x+(b^2-n) => P(a,b,n) in order to conclude that P(b,nb-a,n)

n/t Hey you two, merry Xmas! Thank you for your comments. Maybe Rob was talking about: "Since alpha=nb-a is also a root, P(b,nb-a,n)." If I used <=> there, this step might make more sense. I think it's more a matter of style than correctness. Changing to <=> or not is okay with me. :) Hoi Pang: I don't live in HK. But I will sure wanna talk with you on other topics thru emails. I will email soon, feel free to email me back. :)

哈？ 原本您不在香港？您還說沒錢去非洲，只能去長洲．哈哈！ 您現在在哪一洲？長洲還是短洲？ 獎品怎樣交到您手裏？這樣吧，您出來回的機票和住宿費，我親自把獎品交給您，好嗎？

海澎: 我會回香港嘛。但不會到非洲。儲定錢先﹐儲夠就去北非。:)

SC: It's only a minor technicality really. Anyway your proof is nice.

A simple proof Here we are going to use indirect proof. Suppose that a != b (“a!=b” means a is not equal to b), and (a^2+b^2)/(ab+1)=k ……(1), where k is a positive integer and is not a perfect square. Therefore, we have an indefinite equation, a^2-kab+b^2=k ……(2) It is clear that there exists a solution (a,b) such that ab>0. (since if ab<0, then –ab>1, therfor a^2+b^2<=0, impossible!) (<= means less than or equal to) Now we can choose a pair of solution (a,b) s.t a>0,b>0 and a+b is MIMINUM. And suppose that a>=b and k and b are fixed, so (2) is now a quadratic equation in a, Its roots are, say, a and a’. We have, a+a’=kb ……(3) aa’=b^2-k ……(4) From (3) and (4), we know that a’ is an integer and a’ !=0, otherwise k is a prefect square which is contradictory to the assumption. Since a>0, so a’>0. And from 4 a’=(b^2-k)/a <=(b^2-1)/a <=(a^2-1)/a < a As (a’,b) is a solution of (2)and a’>0,b>0, but a’+b < a+b which is contradictory to our selection of (a,b). (since a+b is minimum!) Therefore is k is a prefect square!

Great !

請教 testing > It is clear that there exists a solution (a,b) such that ab>0. < 想知如果 a, b 是互質的, 而且 a != b 的話, 那這個solution(a,b) 還存在嗎?

JoeJoneS 你沒看清楚題目. 題目是說: *如果* a^2+b^2能被1+ab整除的話,那麼其商是一個完全平方數. 如果 a^2+b^2不能被1+ab整除,其商當然不是一個完全平方數,但這與題目的條件無關.

我想知的是 "a^2+b^2不能被1+ab整除" 會否得出a,b 是互質? 我只知當 a,b 是互質和不相等時, 原題目是無解的

JoeJoneS 你的意思是: 1. 當 a,b 是互質和不相等時,a^2+b^2不能被1+ab整除, 還是 2.當 a,b 是互質和不相等且a^2+b^2能被1+ab整除時,其商不是一個完全平方數? 如果你的意思是1,即使是真的,那又如何? 如果你的意思是2,請證明,或舉一例.

JoeJoneS >>想知如果 a, b 是互質的, 而且 a != b 的話, 那這個solution(a,b) 還存在嗎?<< I have no idea on your conjecture. I can't prove or disprove it! BTW, you may ask a wrong question. I think you are going to ask: If there exists integers a,b s.t. 1+ab|(a^2+b^2) and that quotent is a prefect square, then gcd(a,b)!=1 .