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Congruum Problem


Date: 04/04/2002 at 18:06:35
From: Dr. Allan
Subject: Congruum problem

In researching suitable problems for my students I have found a 
reference to Fibonacci and his congruum problem. Something 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?

Sincerely,
Dr. Allan


Date: 04/05/2002 at 16:04:44
From: Dr. Paul
Subject: Congruum problem

I went to the library this morning and picked up a copy of Ore's 
_Number Theory and its History_. The solution to the congruum problem 
is listed on pages 188-193 and I found it very interesting reading.

What follows is essentially Ore's solution with my commentary added 
where I felt Ore's details were (probably intentionally) lacking 
sufficient explanation:

We want to find a number x such that simultaneously:

   x^2 + h = a^2,       x^2 - h = b^2  (equation 8-16)

and determine for which h rational solutions x can exist. We shall 
first determine the solutions in integers, and this depends, as we 
shall see again, on the Pythagorean triangle. When the second equation 
in (8-16) above is subtracted from the first, one has:

   2*h = (a^2 - b^2) = (a-b)*(a+b)    (8-17)

Since the left-hand side is even, a and b must both be odd or both be 
even. Therefore, a - b is even. Hence

   a - b = 2*k

and k must be a divisor of h since according to (8-17) we have:

   2*h = 2*k*(a+b)

or

   h = k*(a+b)

which implies k divides h.

It follows that 

   (a + b) = h/k

and by adding and subtracting the last two equations, one finds

   a = h/(2*k) + k,   b = h/(2*k) - k

When these two expressions are substituted into the original equations 
(8-16), we obtain:

   x^2 + h = [h/(2*k) + k]^2 = [h/(2*k)]^2 + h + k^2

and

   x^2 - h = [h/(2*k) - k]^2 = [h/(2*k)]^2 - h + k^2

Adding these equations and removing a common factor of two gives:

   x^2 = [h/(2*k)]^2 + k^2

In our case of interest, we have:

   x = 15
   h = 216
   a = 21
   b = 3
   k = 9

Therefore, the three numbers 

   x, h/(2*k), k

form a Pythagorean triangle. Determine the lesser of h/(2*k) and k (in 
this case k = 9 < 12 = 216/18 = h/(2*k)) so that we can write:

   x = t*(m^2 + n^2)

   k = t*(m^2 - n^2)                 (**)

   h/(2*k) = 2*m*n*t

where t is some integer and the expressions in m and n define a 
primitive solution of the triangle.

In actuality, we have:

   x = 15
   k = 9
   h = 216

so clearly, t = 3 and this forces m = 2, n = 1

As an aside, if h/(2*k) > k then we pick

   h/(2*k) = t*(m^2 - n^2)

   k = 2*m*n*t

Now we take the product of the last two expressions in (**) above to 
obtain as the general solution to (8-16)

   x = t*(m^2 + n^2),      h = 4*m*n*(m^2 - n^2)*t^2  (8-18)

Now we make a slight reduction to this solution. Let us suppose that 
we have a solution x of (8-16), where x has the factor t and h at the 
same time has the factor t^2 (this is the case in our example since 
x = 15 is divisible by three and h = 216 is divisible by 9). Then we 
can write:

   x = x_1 * t,        h = h_1 * t^2

From (8-16), we obtain:

   [(x_1)^2 * t^2] + (h_1 * t^2) = a^2 

and

   [(x_1)^2 * t^2] - (h_1 * t^2) = b^2

Then a^2 and b^2 are both divisible by t^2 so a and b are both 
divisible by t:

   a = a_1 * t,    b = b_1 * t

This gives:

   [(x_1)^2 * t^2] + (h_1 * t^2) = [(a_1)^2 * t^2]

and

   [(x_1)^2 * t^2] - (h_1 * t^2) = [(b_1)^2 * t^2].

Cancelling the factor of t^2 gives:

   (x_1)^2 + h_1 = (a_1)^2

and

   (x_1)^2 - h_1 = (b_1)^2

When no further reduction is possible, we shall say that we have a 
primitive solution. When this reduction is applied to (8-18), we 
obtain:

   x = m^2 + n^2

   h = 4*m*n*(m^2 - n^2)

The numbers m and n now produce a primitive Pythagorean triple.

Thus your solution of

   15^2 + 216 = 21^2

   15^2 - 216 = 3^2

Is really just a multiple of the primitive solution:

   5^2 + 24 = 7^2

   5^2 - 24 = 1^2

This corresponds to the first row in the table here:

   Congruum Problem - MathWorld - Eric Weisstein
   http://mathworld.wolfram.com/CongruumProblem.html   

i.e., take m = 2, n = 1.  Then x = 5, h = 24, a = 7, b = 1

and as with Pythagorean triples, letting t be a natural number gives 
an infinite number of solutions.

Neat problem!

I hope this is clear - let me know if I'm skipping steps where I need 
to be filling in the details.

Finally, notice that writing

   h = 4*m*n*(m-n)*(m+n)

is only possible when h is an element of a primitive congruum. That's 
why your attempts failed. You needed to write

   h = 4*m*n*(m-n)*(m+n)*t^2

which works easily when h = 216, m = 2, n = 1, t = 3.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/   


    
Associated Topics:
College Number Theory
High School Fibonacci Sequence/Golden Ratio
High School Number Theory

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