一道IMO题目

wayne · 发表于 4 天前

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1988年IMO一道题,据说陶哲轩没做出来.
1) 原题: 如果$a,b,y$是正整数,$y=\frac{a^2+b^2}{1+ab}$, 求证y是平方数.
2) 我的问题: 如果$a,b$都是有理数,$y=\frac{a^2+b^2}{1+ab}$, y可以是哪些正整数呢

wayne · 发表于 4 天前

第二问.可能OEIS没有收录
平方数的时候,解是$[a,b,y]=[\frac{t (2 U-t^2)}{U^2-t^2 U+1},\frac{t (U^2-1)}{U^2-t^2 U+1},t^2]$
非平方数的话,10000以内,下面这些数都可以,但不知道是什么规律

10,20,34,52,65,73,74,130,148,160,164,202,226,241,244,265,281,290,340,394,416,436,450,452,505,514,569,577,580,586,601,641,650,720,724,745,801,802,820,848,865,884,898,916,929,970,976,1044,1060,1073,1098,1105,1152,1154,1226,1252,1280,1305,1321,1345,1348,1354,1360,1396,1460,1546,1570,1585,1602,1604,1609,1620,1665,1684,1696,1721,1777,1780,1801,1802,1856,1865,1873,1924,2009,2020,2036,2041,2050,2080,2089,2164,2176,2180,2250,2305,2306,2314,2340,2404,2425,2448,2452,2473,2529,2545,2594,2626,2644,2720,2740,2785,2801,2848,2880,2890,2900,2929,2977,3044,3089,3121,3145,3152,3177,3202,3298,3305,3321,3361,3385,3466,3529,3530,3604,3636,3649,3673,3700,3744,3748,3769,3796,3825,3856,3865,3874,3881,3889,3970,3985,4034,4036,4052,4129,4176,4196,4234,4273,4304,4321,4384,4385,4420,4426,4441,4546,4561,4610,4705,4777,4801,4804,4810,4852,4880,4948,4969,5002,5044,5105,5140,5186,5200,5204,5209,5281,5364,5386,5410,5440,5473,5545,5584,5602,5620,5636,5641,5652,5706,5716,5760,5785,5825,5834,5905,5924,6052,6100,6145,6148,6154,6176,6201,6273,6274,6304,6352,6370,6401,6500,6505,6516,6544,6577,6626,6649,6660,6673,6689,6705,6730,6761,6841,6948,6961,6964,7060,7081,7114,7120,7121,7129,7156,7177,7202,7209,7234,7300,7380,7444,7456,7465,7489,7540,7561,7585,7650,7652,7690,7696,7760,7769,7780,7825,7840,7940,7969,7985,8066,8080,8116,8145,8161,8185,8194,8224,8226,8273,8320,8336,8345,8352,8425,8452,8521,8548,8665,8674,8676,8689,8705,8714,8770,8820,8865,8884,8948,8980,9001,9034,9040,9090,9124,9169,9225,9250,9281,9316,9346,9524,9529,9556,9610,9745,9760,9769,9802,9841,9866,9929

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mathe · 发表于 4 天前

y可以写成两个正整数平方和形式，而且这两个正整数乘积加1是完全平方数

wayne · 发表于 4 天前

mathe 发表于 2025-8-4 22:56
y可以写成两个正整数平方和形式，而且这两个正整数乘积加1是完全平方数

很不错. 这代表了一大类. 不过,我比对了一下数据,发现漏解也挺多的.漏掉的如下:

241,569,586,641,745,865,898,929,976,1152,1280,1305,1321,1348,1585,1602,1620,1665,1696,1856,1865,1873,2009,2020,2306,2404,2448,2529,2545,2644,2720,2785,2801,2880,2929,2977,3121,3321,3385,3649,3673,3744,3748,3769,3865,3881,3889,3985,4034,4129,4176,4196,4273,4304,4426,4441,4546,4561,4705,4777,4801,4852,4969,5105,5186,5281,5364,5386,5440,5584,5602,5636,5641,5652,5716,5760,5905,5924,6145,6201,6304,6352,6370,6516,6577,6626,6660,6673,6689,6961,7081,7114,7120,7121,7129,7177,7209,7465,7489,7561,7585,7652,7696,7760,7825,7840,7969,7985,8066,8080,8145,8161,8224,8226,8320,8336,8425,8521,8548,8665,8674,8676,8689,8705,8820,8865,8884,9001,9034,9090,9124,9346,9529,9556,9610,9760,9769,9866,9929

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比如$y=241$的时候,存在参数解 $[a,b]=[\frac{U^2-610 U+73504}{9 (U^2-241 U+1)},\frac{-64 U^2-2 U+305}{9 (U^2-241 U+1)}]$, 但是$241=15^2+4^2, 15*4+1=61$
比如$y=569$的时候,存在参数解 $[a,b]=[\frac{-5 U^2-278 U+79096}{27 \left(U^2-569 U+1\right)},\frac{-2984 U^2+10 U+139}{27 \left(U^2-569 U+1\right)}]$, 但是$569=20^2+13^2, 20*13+1=261$

mathe · 发表于 3 天前

我们设a,b公分母为u, 于是$a=\frac m u, b=\frac n u$得
$y=\frac{m^2+n^2}{u^2+mn}$
或者我们可以写成$m^2-ymn+n^2-yu^2=0$
对于给定的y,如果存在整数m,n满足这个方程，那么其中必有有$\max\{|m|,|n|\}$最小的一组解。
由于我们把上面看成m的方程后，必然有两个整数解，满足$|m_1m_2|=|n^2-yu^2|$,所以如果最小解满足$|m|\ge |n|$，
那么只能$|n|^2 \le |m|^2\le yu^2$, 由于$n^2-yu^2\le 0$,我们可以知道方程还有一组解其中$m,n$符号不同，于是我们可以改写成
$y=\frac{m^2+n^2}{u^2-mn}$
其中特别的$y=m^2+n^2, u^2-mn=1$就是我前面提到的情况。

wayne · 发表于 3 天前

对于 $a^2 + b^2 - (1 + a b) y=0$,关于b的二次方程,判别式是$a^2 y^2-4 (a^2- y) =t^2$ ,此二次曲线的参数解是$[t,y]=[-\frac{a (a^2 U^2-4 a^2+4 U)}{2 (a^2 U+2)},\frac{a^2 (U^2+4\right)}{2 (a^2 U+2)}]$
所以y的表达式$y=\frac{a^2 (U^2+4\right)}{2 (a^2 U+2)}$,对应的b是$b_1=\frac{a U}{2},b_2=\frac{a (2 a^2-U)}{a^2 U+2}$

也就是$\{a,U\}\to {\frac{m}{n},\frac{p}{q}},$得到 $y=\frac{m^2 (p^2+4 q^2)}{2 q (m^2 p+2 n^2 q)}$

mathe · 发表于 3 天前

我们看$m^2+n^2=y(u^2-mn)$
设$(m,n)=d, m=dm_0,n=dn_0, gcd(u,d)=1$
我们得到方程
$d^2(m_0^2+n_0^2)=y(u^2-d^2m_0n_0)$
由于gcd(u,d)=1,所以必然有$d^2|y$,由此得到$u^2-d^2m_0n_0|m_0^2+n_0^2$，于是对于每个$m_0^2+n_0^2$的因子s,可以对Pell方程$u^2-d^2m_0n_0=s$进行求解
对于满足这条件的$(m_0,n_0,u,d)$,可以有$y=\frac{d^2(m_0^2+n_0^2)}{u^2-d^2m_0n_0}$. 不过这种方式还是很难进行枚举。
而特别对于d=1时，我们会要求$u^2-mn|m^2+n^2$的情况即可满足要求，这时$y=\frac{m^2+n^2}{u^2-mn}$, 对应y的所有非平方素因子也都是模4余1的情况。
比如241对应m=64,n=1,u=9.

northwolves · 发表于 3 天前

试了几个数字，发现若$c=\frac{a^n+b^n}{(a b)^{n-1}+1}$是一个整数，似乎 c 也是某个整数d的n次幂。

wayne · 发表于 3 天前

mathe 发表于 2025-8-5 08:27
我们设a,b公分母为u, 于是$a=\frac m u, b=\frac n u$得
$y=\frac{m^2+n^2}{u^2+mn}$
或者我们可以写成 ...

我发现给定$u>1$的任何正整数,总有解,不过解的个数是有限的. 简单跑了下代码, 答案如下 ${y,m/u,n/u}$

{10,-(1/2),3/2}
{20,-(2/3),4/3}
{65,-(1/3),8/3}
{34,-(3/4),5/4}
{226,-(1/4),15/4}
{52,-(4/5),6/5}
{73,-(3/5),8/5}
{148,-(2/5),12/5}
{265,-(1/5),23/5}
{577,-(1/5),24/5}
{74,-(5/6),7/6}
{1226,-(1/6),35/6}
{160,-(4/7),12/7}
{580,-(2/7),24/7}
{1105,-(1/7),47/7}
{2305,-(1/7),48/7}
{130,-(7/8),9/8}
{450,-(3/8),21/8}
{3970,-(1/8),63/8}
{164,-(8/9),10/9}
{281,-(5/9),16/9}
{416,-(4/9),20/9}
{1604,-(2/9),40/9}
{241,-(1/9),64/9}
{3121,-(1/9),79/9}
{6401,-(1/9),80/9}
{202,-(9/10),11/10}
{1098,-(3/10),33/10}
{9802,-(1/10),99/10}
{244,-(10/11),12/11}
{436,-(6/11),20/11}
{601,-(5/11),24/11}
{916,-(4/11),30/11}
{1609,-(3/11),40/11}
{3604,-(2/11),60/11}
{7081,-(1/11),119/11}
{14401,-(1/11),120/11}
{290,-(11/12),13/12}
{20450,-(1/12),143/12}
{340,-(12/13),14/13}
{505,-(8/13),21/13}
{820,-(6/13),28/13}
{1780,-(4/13),42/13}
{3145,-(3/13),56/13}
{7060,-(2/13),84/13}
{13945,-(1/13),167/13}
{28225,-(1/13),168/13}
{394,-(13/14),15/14}
{1546,-(5/14),39/14}
{4234,-(3/14),65/14}
{586,-(1/14),155/14}
{38026,-(1/14),195/14}
{452,-(14/15),16/15}
{848,-(8/15),28/15}
{1073,-(7/15),32/15}
{3152,-(4/15),56/15}
{12548,-(2/15),112/15}
{2545,-(1/15),208/15}
{24865,-(1/15),223/15}
{50177,-(1/15),224/15}
{514,-(15/16),17/16}
{2626,-(5/16),51/16}
{7234,-(3/16),85/16}
{65026,-(1/16),255/16}
{720,-(12/17),24/17}
{1360,-(8/17),36/17}
{865,-(7/17),41/17}
{2340,-(6/17),48/17}
{5200,-(4/17),72/17}
{9225,-(3/17),96/17}
{20740,-(2/17),144/17}
{41185,-(1/17),287/17}
{82945,-(1/17),288/17}
{650,-(17/18),19/18}
{104330,-(1/18),323/18}
{724,-(18/19),20/19}
{801,-(15/19),24/19}
{1044,-(12/19),30/19}
{1396,-(10/19),36/19}
{2089,-(8/19),45/19}
{3636,-(6/19),60/19}
{5209,-(5/19),72/19}
{8116,-(4/19),90/19}
{14409,-(3/19),120/19}
{32404,-(2/19),180/19}
{6961,-(1/19),344/19}
{64441,-(1/19),359/19}
{129601,-(1/19),360/19}
{802,-(19/20),21/20}
{3298,-(7/20),57/20}
{17698,-(3/20),133/20}
{159202,-(1/20),399/20}
{884,-(20/21),22/21}
{1721,-(11/21),40/21}
{2036,-(10/21),44/21}
{3089,-(8/21),55/21}
{7769,-(5/21),88/21}
{12116,-(4/21),110/21}
{2644,-(2/21),212/21}
{48404,-(2/21),220/21}
{96361,-(1/21),439/21}
{193601,-(1/21),440/21}
{970,-(21/22),23/22}
{4810,-(7/22),69/22}
{25930,-(3/22),161/22}
{2314,-(1/22),411/22}
{233290,-(1/22),483/22}
{1060,-(22/23),24/23}
{1345,-(16/23),33/23}
{2080,-(12/23),44/23}
{2425,-(11/23),48/23}
{4420,-(8/23),66/23}
{7780,-(6/23),88/23}
{17440,-(4/23),132/23}
{30985,-(3/23),176/23}
{69700,-(2/23),264/23}
{138865,-(1/23),527/23}
{278785,-(1/23),528/23}
{1154,-(23/24),25/24}
{13250,-(5/24),115/24}
{330626,-(1/24),575/24}
{1252,-(24/25),26/25}
{1777,-(16/25),39/25}
{2473,-(13/25),48/25}
{2848,-(12/25),52/25}
{6148,-(8/25),78/25}
{3985,-(7/25),89/25}
{10852,-(6/25),104/25}
{24352,-(4/25),156/25}
{43273,-(3/25),208/25}
{97348,-(2/25),312/25}
{10273,-(1/25),591/25}
{21745,-(1/25),608/25}
{194065,-(1/25),623/25}
{389377,-(1/25),624/25}
{1354,-(25/26),27/26}
{2250,-(15/26),45/26}
{5706,-(9/26),75/26}
{18250,-(5/26),135/26}
{50634,-(3/26),225/26}
{25546,-(1/26),659/26}
{455626,-(1/26),675/26}
{1460,-(26/27),28/27}
{2900,-(14/27),52/27}
{3305,-(13/27),56/27}
{8345,-(8/27),91/27}
{10865,-(7/27),104/27}
{569,-(5/27),139/27}
{33140,-(4/27),182/27}
{132500,-(2/27),364/27}
{5105,-(1/27),647/27}
{11545,-(1/27),688/27}
{264265,-(1/27),727/27}
{529985,-(1/27),728/27}

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wayne · 发表于前天 16:13

计算了10000项, 提交了一个数列.https://oeis.org/A386855 positive non-square integers of the form (a^2+b^2)/(1+ab) for rational numbers a and b.

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