Problems & Puzzles:
Solutions of A2
+X2 = Y2, A = odd
Isaac Rojas, from Costa Rica, noticed (13/06/01) than in the
+X2 = Y2
if A is an odd prime then there is only one solution-pair (x, y).
Question: What is the general expression that
gives the quantity
of solutions-pairs (X, Y) of A2
+X2 = Y2
If A= 3
the the unique solution is (x, y) = (4, 5)
if A=5 then the unique solution is (x, y) = (12, 13)
But if A = 9 then the solutions are two: (x, y) = (12,
15) & (40, 41)
if A is odd but not necessarily prime.
Several puzzlers attempted the solution of this puzzle,
but only Chris Nash got (23/7/01) the complete one:
" ...We seek
the number of solutions to A^2+X^2=Y^2 where A is
odd. We note A^2=Y^2-X^2=(Y-X)(Y+X) In other words, every solution X,Y
with Y>X>0 gives a factorization of A^2 into two distinct integers.
Also the reverse applies - if A^2 is factorized into two distinct
integers, PQ, (with P>Q) then both P and Q are odd, and a solution for
X and Y exists - X = (P-Q)/2, Y=(P+Q)/2.
Hence the result we seek is the number of ways to factor A^2 into
distinct integers. We recall the result:
If X=p1^r1.p2^r2.....pn^rn is the prime factorization of X, then X
has precisely (r1+1)(r2+1)....(rn+1) distinct factors including itself and
We find the prime factorization of A, let us call it
p1^r1.p2^r2.....pn ^rn. Then A^2 has exactly (2.r1+1)(2.r2+1).....(2.rn+1)
distinct factors. This is an odd number, and all of them can be paired to
form two numbers whose product is A^2, except for A itself. Hence the
number of solutions is exactly ((2.r1+1)(2.r2+1).....(2.rn+1) - 1)/2 where
the r_i are the powers of the primes in the factorization of *A*.
For example, if A is prime, then r1=1, and the solution count is
indeed 1. For A=9, the factorization has r1=2, and so the number of
solutions is ((2*2+1)-1)/2=2. As another example, if A is the product of
the first 100 primes, then the number of solutions is exactly (3^100-1)/2."
Partial solutions came from Jud McCranie, Jean Brette, Ole Jan, Said
El Aidi and Isaac Rojas - of course.