两上数的平方和

宋开福

（中国湖北宜昌夷陵区东湖高中 邮编443100 邮箱"skf08"skf08@sina.com）

关键词 二次同余方程;两上数的平方和.

摘 要 本文介绍利如何把一个数表作两个数的平方和的一般方法。

一、引言

我们已知道哪些数可以表作两个数的平方和，哪些数不能，对可表作两个数的平方和的数，当这个数较小时，凭观察可直接写出，当这个数较大时，凭观察却不是一件容易的事，下面给出如何由一数来表作两个数的平方和的一般方法.

二、两上数的平方和

把一个数表作两个数的平方和，关键是把质因数4n+1如何分成两个数的平方和，其方法可按以下步骤进行.

（i）解方程x²≡-1（mod p）.∵p=4n+1=>（-1）^(p-1)/2=（-1）²ⁿ=1.

∴方程x²≡-1（mod p）有解，由前面介绍的求二次同余方程的方法求出它的解。x²≡±x₁（mod p）.

（ii）令x₁=n，对于方程（nx）²≡y²（mod p）=> （nx+y）（nx-y）≡0（mod p）=>

nx+y=kp=>kp-nx=y,或nx-y=tp=>kp-nx=y,则

nx≡pk（mod y），令p=nq₁+r₁，0< r₁<n=>

nx≡r₁k（mod y），令n= r₁q₂+r₂，0< r₂< r₁=>

r₂x≡r₁k（mod y），令r₁= r₂q₃+r₃，0< r₃< r₂=>…=> r_nx≡r_n-1k（mod y）.

这里r_n-1，r_n都小于r，r是根号下P再加1的高斯函数，但r_n-2>r. 于是取x= r_n，y= r_n-1则有

（nx+y）（nx-y）≡0（mod p）=> （nx）²≡y² （mod p），又x，y<r ,r为根号p再加1的高斯函数，

又n²≡-1（mod p），=> x² +y²≡p.

三、应用举例

例1 把233表成两个数的平方和.

解（i）解方程x²≡-1（mod 233）.

∵2n+1=233=>n=116=>-n/2=-58 . ∴x²≡-58（mod 233）是方程x≡±116（mod 233）的基准方程，

在(233k+57)/m(m+1)中，k_个位=1,3,5，则k=3.

当k=3时，233k+57=756=27×28=>116-27=89.

∴x²≡-1（mod 233）的解是x≡±89（mod 233）.

（ii）令（89x）²≡y² （mod 233）=> 89x≡233t（mod y）=>89x≡55t（mod y）=>

34x≡55t（mod y）=>34x≡21t（mod y）=> 13x≡21t（mod y）=>13x≡8t（mod y）.

∵r为根号下233再加1的高斯函数，则r=16，于是取x=13，y=8，则233= 13² +8².

可见这些数学问题本身确实存在奇妙之处，之所以认为它玄奥，是我们还没发现而已。

参考文献

（Ⅰ）宋开福。初等数论。中国北京、中国戏剧出版社 2007

The square of the number two and

SONG Kai-fu

(China Yiling District Yichang, Hubei 443100 Email East Lake High School Zip Code "skf08" skf08@sina.com)

Key words quadratic congruence equation; two and the number of square.

Abstract This article describes how to benefit a few tables for two and the square of the number of the general approach.

I. INTRODUCTION

We have to know what could be a few tables for two and the square, which can not be on the table for two to the square of the number and the number, when the number of small, with the observation of direct write, when the number of larger when their observation is not an easy task, given how the following few years by a number of tables for the square and the two general methods.

Second, the two on the square of the number and

A few tables for two and the square of the number, the key is the quality factor of 4n +1 how the square of the number two and that its methods can be carried out the following steps.

(I) solution of equation x2 ≡ -1 (mod p). ∵ p = 4n +1 =&gt; (-1) (p-1) / 2 = (-1) 2n = 1.

∴ equation x2 ≡ -1 (mod p) to solve, by the order described above with the quadratic equation method to derive more than its solution. x2 ≡ ± x1 (mod p).

(Ii) so that x1 = n, the equation (nx) 2 ≡ y2 (mod p) =&gt; (nx + y) (nx-y) ≡ 0 (mod p) =&gt;

nx + y = kp =&gt; kp-nx = y, or nx-y = tp =&gt; kp-nx = y, then

nx ≡ pk (mod y), so p = nq1 + r1, 0 &lt;r1 <n=>

nx ≡ r1k (mod y), so n = r1q2 + r2, 0 &lt;r2 &lt;r1 =&gt;

r2x ≡ r1k (mod y), so that r1 = r2q3 + r3, 0 &lt;r3 &lt;r2 =&gt; ... =&gt; rnx ≡ rn-1k (mod y).

Here rn-1, rn are less than r, r is the root of P together with the No. 1 Gaussian function, but rn-2&gt; r. Thus taking x = rn, y = rn-1 are

(Nx + y) (nx-y) ≡ 0 (mod p) =&gt; (nx) 2 ≡ y2 (mod p), then x, y <r ,r under No. p for the root of the Gaussian function combined with 1,

And n2 ≡ -1 (mod p), =&gt; x2 + y2 ≡ p.

Third, the application example

233 cases of Table 1 into two and the square of the number.

Solution (i) solution of equation x2 ≡ -1 (mod 233).

∵ 2n +1 = 233 =&gt; n = 116 =&gt;-n / 2 =- 58. ∴ x2 ≡ -58 (mod 233) is the equation x ≡ ± 116 (mod 233) of the benchmark equation,

In (233k +57) / m (m +1) in, k = 1,3,5 bits, then k = 3.

When k = 3 when, 233k +57 = 756 = 27 × 28 =&gt; 116-27 = 89.

∴ x2 ≡ -1 (mod 233) is the solution x ≡ ± 89 (mod 233).

(Ii) the (89x) 2 ≡ y2 (mod 233) =&gt; 89x ≡ 233t (mod y) =&gt; 89x ≡ 55t (mod y) =&gt;

34x ≡ 55t (mod y) =&gt; 34x ≡ 21t (mod y) =&gt; 13x ≡ 21t (mod y) =&gt; 13x ≡ 8t (mod y).

∵ r under No. 233 for the root of the Gaussian function combined with 1, then r = 16, then take x = 13, y = 8, then 233 = 132 +82.

Apparently, the mathematical problem itself, the existence of magic, mysterious reason that it is we have not found it.

References

(Ⅰ) SONG Kai-fu. Elementary number theory. Beijing, China, the Chinese Drama Publishing House, 2007