In 1988, as a new headquarters for the American Central Intelligence Agency (CIA) was being built in Langley, Virgina, sculptor Jim Sanborn was commissioned to create artwork for the courtyard of the new building.
In 1999, a computer scientist named Jim Gillogly announced that he had solved most of the puzzle. There were four distinct parts to the code, and he had managed to solve the first three.
After his announcement, the CIA revealed that they had actually solved the first three parts internally, the year before. (Later, the U.S. National Security Administration (NSA) also claimed to have solved the first three parts in 1992.)
The fourth section of the code remained unsolved, and to date, no one has brought forth a credible solution for part 4. The code has proven so difficult that its creator, cryptographer and sculptor Jim Sanborn, has provided a tantalizing clue to the New York Times: Characters 64 through 69, the letters N-Y-P-V-T-T, are decoded as B-E-R-L-I-N.
When commenting in 2006 about his error in section 2, Sanborn said that the answers to the first three sections contain clues to the fourth section. In November 2010, Sanborn released a clue, publicly stating that “NYPVTT”, the 64th-69th letters in part four, become “BERLIN” after decryption.
Sanborn gave The New York Times another clue in November 2014: the letters “MZFPK”, the 70th-74th letters in part four, become “CLOCK” after decryption. The 74th letter is K in both the plaintext and ciphertext, meaning that it is possible for a character to encrypt to itself. This means it does not have a weakness, where a character could never be encrypted as itself, that was known to be inherent in the German Enigma machine. It is believed that the “BERLINCLOCK” plaintext may be a direct reference to the Berlin Clock.
Sanborn further stated that in order to solve section 4, “You’d better delve into that particular clock,” but added, “There are several really interesting clocks in Berlin.”[
So what if NSA’s cryptanalysts have also spent their lunch hours—aided by classified code-breaking techniques, massive computing power, and giant stores of data—cracking some of the world’s other great, unsolved cryptographs?
HAWAII + IDAHO + IOWA + OHIO == STATES
Translation of the allegory designed by the Ikhwān al-Ṣafā’ to explain the calculation of the Great Year.
Know further, my brother, that the wise men of India have formulated an allegory of the revolutions of these stars around the Earth, so that the comprehension of this may be brought closer to those who learn and that its representation may be made easier to those who meditate. According to their report, a king among kings built a city with a circumference of 60 parasangs and sent seven individuals to revolve around it at different paces: one of them 1 parasang every day, another one 2 parasangs every day, the third 3 parasangs every day, the
fourth 4 parasangs every day, the fifth 5 parasangs every day, the sixth 6 parasangs every day, and the seventh, 7 parasangs every day. And the king told them: ‘Revolve around this city, starting from this door; when, by the number of your revolutions, you join together at the same door, come [to me] and announce to me how many revolutions each one you will have completed’. Whoever has understood the computation of the revolutions of these individuals around the city and succeeded to figure it by imagination is likely to understand the revolutions of these stars around the Earth, and to figure out how many revolutions it will take them to join together in the first [minute] of Aries, in the same way as they were when they started.
As for the computation [concerning] these individuals, it is as follows. After 60 days, 6 individuals join together at the door of the city. The first one has completed 1 revolution, the second 2 revolutions, the third 3 revolutions, the fourth 4 revolutions, the fifth 5 revolutions, the sixth 6 revolutions. As for the one who completes, every day, 7 [parasangs], he has completed 8 revolutions plus 4/7 of the parasangs of a revolution, so that these individuals must renew the revolution. After 120 days, they join together once again at the door
and each one has completed his first count once again, but the seventh has completed 17 revolutions plus 1/7 of the parasangs [of a revolution], so that they must renew the revolution. After 180 days, the six join together for the third time and each one has completed, for the third time, his first count, but the seventh companion has completed 25 revolutions plus 5/7, so that they must renew the revolution. After 240 days, they join together for the fourth time and each one of them has completed his first count, but the seventh companion has completed 34 revolutions plus 2/7, so that they must renew the revolution. After 300 days, they join
together for the fifth time, but the seventh companion has completed 42 revolutions plus 6/7, so that they must renew the revolution. After 360 days, they join together for the sixth time and each one of them has completed his first count for the sixth time, but the seventh companion has completed 51 revolutions plus 3/7 of the parasangs [of the revolution], so that they must renew the revolution. After 420 days, all of them join together at the door of the city: the first one has completed 7 revolutions, the second 14 revolutions, the third 21 revolutions, the fourth 28 revolutions, the fifth 35 revolutions, the sixth 42 revolutions, and the seventh has completed 60 revolutions. This is the allegory that the wise men of India have invented in order to account for the revolutions of the spheres and the stars around the Earth. Thus, the Earth is like this city that was built with a circumference of 60 parasangs. The seven planets and their revolutions around the Earth are like these seven individuals. The difference in speed and slowness between their movements are like the differences between the courses of these individuals. As for the king, he is the God, the Creator, the Founder – Blessed be the God, Lord of the Worlds.
1. What is the smallest magic square (n) having solution?
2. Get one solution for the next three larger magic squares (n+1, n+2 & n+3)
3. Redo the exercises 1 & 2 with one additional condition: “one of the diagonals should also contain prime numbers only“.