Cayley problem 21, 1997

January 14, 2020

From the Cayley competition, 1997. Consider the equation

$\dfrac{\frac{a}{c}+\frac{a}{b} + 1} {\frac{b}{a} + \frac{b}{c} + 1} = 11$

with the constraint $a + 2b + c \leq 40.$ Find all solutions $a,b,c$ among the natural numbers.

If we set and do a bit of algebra, the variable vanishes and we are left with Organizing the possible values of $a$ , $b$ and $c$ into a table, we get something like…

$a$	$b$	$c$	$a+2b+c$
11	1	1	$11 + 2 + 1 = 14$
11	1	2	$11 + 2 + 2 = 15$
$\cdots$	$\cdots$	$\cdots$	$\cdots$
11	1	27	$11 + 2 + 27 = 40$
$\cdots$	$\cdots$	$\cdots$	$\cdots$
22	2	1	$22 + 4 + 1 = 27$
$\cdots$	$\cdots$	$\cdots$	$\cdots$

If you have good bookkeeping skills, finish the whole table. You should get 42 rows! So there are 42 possible solutions.

But let’s verify this with a little computer programming. Define a predicate that is true when the given conditions of the problem are true:

#lang racket

(define (cayley-21-conditions? a b c)
  (and (= (/ (+ (/ a c) (/ a b) 1)
             (+ (/ b a) (/ b c) 1))
	  11)
       (<= (+ a (* 2 b) c)
           40)))

Then loop through all the possibilities for $a$ , $b$ and $c$ from 1 to 40 and collect them into a giant list of triples $(a,b,c)$ . Then we filter this list by applying the cayley-21-conditions? predicate to each one. Those that pass are kept, those that are false are filtered out.

(define (abc-filter max-a max-b max-c predicate)
  (filter (lambda (args)
            (apply predicate args))
          (for*/list ((a (range 1 (+ max-a 1)))
                      (b (range 1 (+ max-b 1)))
                      (c (range 1 (+ max-c 1))))
            (list a b c))))

Running

> (abc-filter 40 40 40 cayley-21-conditions?)

Yields a list of solutions. Then calling length on this list gives the total number of solutions: 42.

How many solutions would there be if the second condition was $a + 2b + c \leq 100$ ?