密钥总共只有$26^3$种可能，暴力搜索所有组合即得密钥是exp，结果是$129448$。原文如下：
An extract taken from the introduction of one of Euler's most celebrated papers, "De summis serierum reciprocarum" [On the sums of series of reciprocals]: I have recently found, quite unexpectedly, an elegant expression for the entire sum of this series 1 + 1/4 + 1/9 + 1/16 + etc., which depends on the quadrature of the circle, so that if the true sum of this series is obtained, from it at once the quadrature of the circle follows. Namely, I have found that the sum of this series is a sixth part of the square of the perimeter of the circle whose diameter is 1; or by putting the sum of this series equal to s, it has the ratio sqrt(6) multiplied by s to 1 of the perimeter to the diameter. I will soon show that the sum of this series to be approximately 1.644934066842264364; and from multiplying this number by six, and then taking the square root, the number 3.141592653589793238 is indeed produced, which expresses the perimeter of a circle whose diameter is 1. Following again the same steps by which I had arrived at this sum, I have discovered that the sum of the series 1 + 1/16 + 1/81 + 1/256 + 1/625 + etc. also depends on the quadrature of the circle. Namely, the sum of this multiplied by 90 gives the biquadrate (fourth power) of the circumference of the perimeter of a circle whose diameter is 1. And by similar reasoning I have likewise been able to determine the sums of the subsequent series in which the exponents are even numbers.
注：某些欧拉计划的镜像站上似乎还有一段完全不同的、摘自圣经的原文以及对应密钥god，该版本流传的原因不详，此处不作收录。
with open("059_cipher.txt") as f:
for line in f:
s=[int(w) for w in line.split(",")]
for i in range(97,123):
for j in range(97,123):
for k in range(97,123):
p=[i,j,k]
a=0
m=""
t=0
while a < len(s) and len(m)==a:
b=s[a]^p[a%3]
if (b >=97 and b <=122) or (b >=65 and b <=90) or (b >=32 and b <=47) or (chr(b) in "0123456789+,;.?!:()-[]"):
m+=chr(b)
t+=b
a+=1
if len(m)==len(s):
print t,[chr(z) for z in p]
print m
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