update

python/131.md

@@ -2,12 +2,12 @@ $$
 m^3=n^3+n^2p
 $$
 
-let $m=n+k$ , then
+设 $m=n+k$ ，则
 $$
 3n^2k+3nk^2+k^3=n^2p\\
 (p-3k)n^2-3k^2n-k^3=0
 $$
-witch $p>3k$ . solving this function, we get
+其中有 $p>3k$ 。解此方程，得
 $$
 n = \frac{3k^2 \pm k\sqrt{4kp-3k^2}}{p-3k}
 $$

python/137.md

@@ -6,33 +6,33 @@ $$
 xA_F(x)= \; x^2F_1+x^3F_2+x^4F_3+\cdots
 $$
 
-$(1)+(2)$, and consider that $F_1=F_2$, then we have
+(1)+(2)，且考虑到 $F_1=F_2$ ，有
 $$
 (1+x)A_F(x)=xF_2+x^2F_2+x^3F_3+\cdots
 $$
-so
+所以
 $$
 x(1+x)A_F(x)=A_F(x)-xF_1
 $$
-then
+即
 $$
 A_F(x)=\frac{x}{1-x-x^2}
 $$
-we want that $A_F(x)$ could be a nature number, so let
+想要 $A_F(x)$ 是自然数，只需设
 $$
 \frac{x}{1-x-x^2}=n
 $$
-then
+即
 $$
 nx^2+(n+1)x-n=0
 $$
-solve this, get
+解此方程，得
 $$
 x=\frac{\sqrt{5n^2+2n+1}-(n+1)}{2n}
 $$
-only $5n^2+2n+1$ is square number that can make x to be a rational.
+只有 $5n^2+2n+1$ 是平方数时， $x$ 才可能是有理数。
 
-let's try to analyse how $n$ increasing when $5n^2+2n+1$ is square number.
+尝试分析 $N=5n^2+2n+1$ 是平方数时， $n$ 的变化规律
 
 | n          | increasing rate of n |
 | ---------- | -------------------- |
@@ -49,4 +49,4 @@ let's try to analyse how $n$ increasing when $5n^2+2n+1$ is square number.
 | 507544127  | 6.854101982          |
 | 3478759200 | 6.854101969          |
 
-it is obvious that the rate of $n$ is converge to $6.8541\cdots$ . with using search engein we know $\varphi^4=6.8541\cdots$ and $\varphi=\frac{\sqrt{5}-1}{2}$.
+显然 $n$ 的变化率收敛于 $6.8541\cdots$ 。通过搜索引擎可知 $\varphi^4=6.8541\cdots$ 其中 $\varphi=\frac{\sqrt{5}-1}{2}$ 。

python/tools/algebra.py

@@ -4,3 +4,9 @@ def factorial(n, multi=1):
         return multi
     else:
         return factorial(n - 1, multi * n)
+
+def is_root(n, exp=2):
+    root = int(n ** (1 / exp))
+    if root ** exp == n:
+        return root
+    return 0

python/tools/number_theory.py

@@ -20,12 +20,11 @@ def pow_mod(base, exp, mod):
 
 
 def make_prime(limit):
-    if limit < 5:
-        if limit < 2: return
-        yield 2
-        if limit < 3: return
-        yield 3
-        return
+    if limit < 2: return
+    yield 2
+    if limit < 3: return
+    yield 3
+    if limit < 5: return
 
     n = (limit + 1) // 6
 
@@ -46,8 +45,6 @@ def make_prime(limit):
             a[f::p] = [False] * ((n - f - 1) // p + 1)
             b[g::p] = [False] * ((n - g - 1) // p + 1)
 
-    yield 2
-    yield 3
     for i in range(n):
         if b[i]:
             yield 6 * i + 5