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:

"[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).

Posted · Report post

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?

## Share this post

## Link to post

## Share on other sites