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.m.../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?

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# Out of respect to Bonanova--Fibonacci

### #1

Posted 02 April 2013 - 10:46 PM

### #2

Posted 04 April 2013 - 04:06 PM

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, 04 April 2013 - 04:10 PM.**

### #3

Posted 10 April 2013 - 07:39 PM

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 4*ab*(*a*+*b*)(*a-b*) is the more fundamental

definition, from which x^{2} + *c* and x^{2} - *c* being squares can be derived.

Putting another way, *x*^{2} + *c* and *x*^{2} - *c* being squares seems to be

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

*Vidi vici veni.*

### #4

Posted 11 April 2013 - 12:46 AM

### #5

Posted 11 April 2013 - 10:02 AM

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.

*Vidi vici veni.*

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