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Fibo was a winner (Posted on 2017-12-17) Difficulty: 4 of 5
Leonardo of Pisa's fame as a gifted problem solver caused the Holy Roman Emperor Frederick II to come to Pisa joined by a team of mathematicians to publicly test his reputation.

One of the challenges was:

Find a square number from which, when five is added or subtracted, always arises a square number.

(Clearly, the quest was for a rational number, since no integer solution exists.)

After pondering for a while the genial Leonardo delivered a correct solution (strictly pen and paper - in 1225 there were neither laptops nor Google) and nowadays there is no exact knowledge how he accomplished the task.

Try to tackle it, using only the simple tools available in the XIII century, and only then allow yourself the indulgence of having a small toolbox of Google and/or software.

Above all – enjoy!

No Solution Yet Submitted by Ady TZIDON    
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Solution | Comment 1 of 4

This problem is actually related to Ady's previous one, any two out of 5. http://perplexus.info/show.php?pid=11104&cid=59243.

With respect to Ady however, we know exactly how Leonardo of Pisa = Fibonacci = Pisano solved the problem because he wrote a book about it, Liber Quadratorum. Strictly speaking the 'challenge' was rigged - the object was to demonstrate Fibonacci's genius and thus enhance the prestige of Frederick II's Court - as it did, Frederick II becoming famed as 'Stupor Mundi' - the Wonder of the World.

Liber Quadratorum is paraphrased here: http://www.jstor.org/stable/2974039?origin=JSTOR-pdf. The relevant proposition is at p6 (Proposition 15, in other editions it is Proposition 16), but the key insight is at Proposition 12. I will come to it in a moment.

Fibonacci almost certainly started with geometric ideas in mind, to do with right-angled triangles.

Let 2b^2=c^2+a^2, and consider small values. The formulation a= 2xy+(y^2-x^2), b=(x^2+y^2),  c=2xy+(x^2-y^2) drops out intuitively on inspection, and we reach the formulations for the  common difference, d:
(2xy+(x^2-y^2))^2-(x^2+y^2)^2 = 4(x^3y-xy^3) = 4xy(x-y)(x+y) = d
(x^2+y^2)^2-(2xy+(y^2-x^2))^2 = 4(x^3y-xy^3) = 4xy(x-y)(x+y) = d

It is then trivial to show that 4xy(x-y)(x+y) has to be divisible by 24. This is what Fibonacci showed in Proposition 12. The numbers that can be d are 'congruum' numbers.

In Proposition 16, Fibonacci starts by 'finding a congruum whose fifth part is a square, such as 720.' He provides a method for computing whole-number solutions which 'added to, or subtracted from, a square number, in either case leaves a square number' in Proposition 13 - essentially the same identities as set out above.

More generally, for there to be a rational solution to b^2-4xy(x-y)(x+y)=a^2, b^2+4xy(x-y)(x+y)= c^2, it will first be necessary both to clear the multiple of 24, and to form a suitable square from the term in x and y, for use as the denominator in the fractional equation. Hence we seek some 4xy(x-y)(x+y) equal to 144a^2P, with P the desired prime.  But if so, then xy^3-x^3y = (6a)^2(2P-1),  assuming that the prime is odd. Small prime solutions are {5,7,41}.

Since 5 is the desired difference, we have 4(2*18^3-2^3*18) = 12^2*8^2*5, giving d=46080. So b^2-46080=a^2, c^2-46080=b^2, producing the integer solution a = 248, b = 328, c = 392. Now the denominator is (12*8), and indeed (328/96)^2-(248/96)^2 = (392/96)^2-(328/96)^2 = 5 . This can be simplified to (49/12)^2 - (41/12)^2  = (41/12)^2 - (31/12)^2 = 5, the solution given by Fibonacci. Using the same method, the corresponding solutions for 7 and 41 are:

x y d  a b c denom. z
9 16 100800  113 337 463 120 7 
16 25 590400  431 881 1169 120 41

(463/120)^2 - (337/120)^2 = (337/120)^2-(113/120)^2 = 7
(1169/120)^2 - (881/120)^2 = (881/120)^2-(431/120)^2 = 41  

Edited on December 17, 2017, 11:57 pm
  Posted by broll on 2017-12-17 23:55:41

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