来来来，做道数学题

[TheMatrix2]

楼上诸位大侠，既然有答案了，能不能给个(1, 0)以外的解示范一下。

除(1,0)以外无解 - 大家都在证明这个问题。

[meiyoumajia]

这题肯定没有简单解法。

it is not too hard to prove that the last two digits of the two sides of
x^3=y^2+1
can only be 1 (or 01), 25, 17, 37, 57, 77, or 97

then it follows that
x must end with 1, 3 or 5 (x-1 has a remainder of 0, 2 or 4 when divided by 10);
and
y must be an even number

By using the equation, we can also find out that the even number x-1 must be a multiple of 4 (because x^2+x+1 is odd, and because y^2 is even)

so
x-1 = 20m + j (j is 0, 12, or 4)

the problem is then reduced to 3 sub-problems.

It is up to you to decide if you are going to spend your time to prove that both j and m must be zeros.

[meiyoumajia]

【我把我之前弄错的内容删除了】

[wsnren]

谢谢。

连分数解Pell方程这里我也没有仔细看，在Wolfram上找到一个表。刚才仔细看了一下，的确是说列出的是最小整数解。无穷多解make sense。

现在
y^2=3x^2+3x+1
有无穷多有理数解。

但是x本身也必须是完全平方数。看来还得用到这一条。这也不容易，因为它没有一个直接的表达式。

是的，得用x是平方这个fact。不过就算y^2=3x^4+3x^2+1的话，又是一条椭圆曲线，比最开始那条还复杂些。

[wsnren]

楼上诸位大侠，既然有答案了，能不能给个(1, 0)以外的解示范一下。

(1,0)是唯一的解。

[FGH]

是否存在（0,1）和（8,9）之外的一对相差为一的乘方数？
乘方数是指一个数的大于一的指数次乘方。比如25,27,32.

[wsnren]

是否存在（0,1）和（8,9）之外的一对相差为一的乘方数？
乘方数是指一个数的大于一的指数次乘方。比如25,27,32.

这是著名的Catalan Conjecture，1844年就存在了。十几年前才彻底搞定。

[TheMatrix2]

这是著名的Catalan Conjecture，1844年就存在了。十几年前才彻底搞定。

哦。搞定是啥结论？没有其他的了？

[meiyoumajia]

【改进了算法】

[wsnren]

哦。搞定是啥结论？没有其他的了？

对，除了(8, 9)外没有别的正整数幂解。

[TheMatrix2]

when j=0, m must be a multiple of 5^n (n can be ANY positive integer), so m=0. therefore, x=1 is a solution.

when j = 12, y = 2s, s^2 must end with 3, which is impossible

so the problem is reduced to the j=4 sub-problem. (if no solution exists for this case, then x=1 is the only solution.)

你前一篇是对的：
x-1 = 20m + j (j is 0, 12, or 4)

但是这一篇好像不对。j=0的情况就不对。

[meiyoumajia]

你前一篇是对的：
x-1 = 20m + j (j is 0, 12, or 4)

但是这一篇好像不对。j=0的情况就不对。

你对。我把那两个情况都弄错了。

[Urus]

尼玛 这不是Mordell's Equation的个例么

前几天的一道作业题，可能得用到超过高中的知识。找出方程

y^2=x^3-1

所有的整数解。

[zzsxt]

找出方程
x^2*|y^2-3*z^2|=y*z
所有的正整数解。

let x=m+1 (不妨设m为>3的正整数)

x^3-1
=(m+1)^3-1
=m*(m^2+3m+3)

分两种情况A和B：

[TheMatrix2]

尼玛 这不是Mordell's Equation的个例么

哦这就是Mordell Equation啊。

网上查了一下，有一个Keith Conrad的notes，讲了各种y^2=x^3+k的情况。奇怪每一个k都是单独处理，没有通用的方法吗？

[meiyoumajia]

[改进了算法】

[meiyoumajia]

肯定没有通用的办法。

对不是有无数个解的情况，都应该能用计算机算出来。

For the previous 3 cases of x:

when j=0,
since x-1= 20m (a multiple of 5), y/2 is a multiple of 5, then m also is a multiple of 5
y is a multiple of 10, and x-1 is a multiple of 100
then the distinctiveness of the two sides' last few digits are very low.

When we first tried to narrow down the ranges of the numbers, we used the knowledge about the last digits of the two sides of the equation.
So further checking more digits/places of either side is not efficient.

One improvement is to introduce a different "basic" numerical system. instead of using the conventional decimal system, we can use, say, a 14-based system.
mathematically/arithmetically, it is equivalent to apply modulo operations over 14, or more.

in the 14-based system, we can check the last few digits.
an alternative way is to check only the last digit, and also check the remainders of modulo over 10, 11, 12, and 13.
For all the three cases, computations show that when the numbers (just mentioned) can possibly match, both m and j have to be zeros.

[meiyoumajia]

Using the new approach, we just need to check 14 pairs of numbers from the two sides of the equation, to see if their remainders match from their being divided by 10, 11, 12, 13, and 14.

So, it is a great improvement.

[TheMatrix2]

肯定没有通用的办法。

对不是有无数个解的情况，都应该能用计算机算出来。

无遗漏的分成有限种情况，然后用计算机检查每一种情况，这也算证明。

[meiyoumajia]

无遗漏的分成有限种情况，然后用计算机检查每一种情况，这也算证明。

有些方程有无数解，因此可能用计算没法解决。