Jump to content
BrainDen.com - Brain Teasers
  • 0

Out of respect to Bonanova--Fibonacci


BMAD
 Share

Question

Recently, Bonanova made a comment to one of my problems about Fibonacci numbers. In all honesty I am not too familiar with their work so I began studying it. In researching it I have found a
reference to their congruum problem. But something about it has me stumped; I hope one or more of you can tell me what I am doing wrong:

Quote from the article at the MacTutor History of Mathematics archive
at St. Andrews University:

http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Fibonacci.html

"[Fibonacci] defined the concept of a congruum, a number of the form
ab(a + b)(a - b), if a + b is even, and 4 times this if a + b is odd
where a and b are integers. Fibonacci proved that a congruum must be
divisible by 24 and he also showed that for x,c such that x^2 + c and
x^2 - c are both squares, then c is a congruum. He also proved that a
square cannot be a congruum."

With x=15 and c=216, we get the two squares 441 and 9, meaning that
216 should be a congruum. Thus we should be able to find numbers a
and b such that

216 = ab(a+b)(a-b) if a+b is even

or

54 = ab(a+b)(a-b) if a+b is odd

Now, b must be smaller than a and d:= ab(a+b)(a-b) is increasing in
both a and b.

Let's look at the possible scenarios.

Say a+b is odd. We want d to equal 54. Our triples (b,a,d) give us

(1,2,6) (1,4,60); thus b=1 is not possible
(2,3,30) (2,5,210); thus b=2 is not possible
(3,4,84); thus b = 3 and b > 3 is not possible (since d is
increasing); thus a+b odd is not possible

Say a+b is even. We want d to equal 216. Our triples (b,a,d) give us

(1,5,120) (1,7,336); thus b=1 is not possible
(2,4,96) (2,6,3849; thus b=2 is not possible
(3,5,360); thus b = 3 and b > 3 is not possible (since d is
increasing); thus a+b even is not possible

So, because 15^2 + 216 = 21^2 and 15^2 - 216 = 3^2 we should have
that 216 is a congruum, but we cannot put it in the form ab(a+b)(a-b).

What is wrong with this argument?

Link to comment
Share on other sites

4 answers to this question

Recommended Posts

  • 0

I think the only flaw is that you assumed that ab(a+b)(a-b) is increasing in a and b. But let us take the derivative of this function with respect to b, keeping 'a' a constant. it is


a^3 - 3(a)(b^2) or a(a^2 - 3(b^2)). this derivative can be negative if a<(3^0.5)b [a is greater than b but may be less than root 3 times b]. This means that this function can be decreasing in b.


I am not quite sure about this because I did not find numbers a and b which give 216. Also, on wikipedia (http://en.wikipedia.org/wiki/Congruum), a list of first few congrua is given which does not contain 216. So, maybe 216 cannot be represented in this way. I did not find any other flaw in your argument.

Edited by unakade
Link to comment
Share on other sites

  • 0

I had read a while back about congrua.

This puzzle gave me a reason to look at them again.

As unakade notes, 216 is not listed by Wolfram as a congruum.

But a sub-multiple (24) is, and it does satisfy both definitions:

25 + 24 = 49
25 - 24 = 1

where 25, 49 and 1 are squares.

24 = 4ab(a+b)(a-b)

where (a,b) = (2,1).

I wasn't able to see where scaled-up versions of a lowest-term congruum

are excluded, but it does seem that 4ab(a+b)(a-b) is the more fundamental

definition, from which x2 + c and x2 - c being squares can be derived.

Putting another way, x2 + c and x2 - c being squares seems to be

a necessary but not sufficient condition for c to be a congruum.

Link to comment
Share on other sites

  • 0

So are we saying that x^2 + C condition is true for all conundrums but not all numbers where this is true for are conundrums? E.g. All squares have four equal sides but not all four equal sided shapes are squares.

I'm saying it seems that is so.

216 is a counter example against the definitions being equivalent.

I read some more about congrua having to correspond to primitive Pythagorean triples,

which includes 24 but excludes 9x24 = 216, but the reason for that requirement escaped me.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...