--- In

primeform@yahoogroups.com,

"Sean A. Irvine" <sairvin@...> wrote:

> Nice! That was a very quick response to the challenge.

Jarek's construction involves this Diophantine problem:

Find a pair of positive coprime integers [a,b], with a<b,

such that there exist two pairs of positive integers,

[c,d] and [e,f], with c < d, c < e < f,

2*a^2 + 2*b^2 = c^2 + d^2 = e^2 + f^2

and distinct positive integers

g = |(4*a*b)^2 -(d^2-c^2)^2|

h = |(4*a*b)^2 -(f^2-e^2)^2|

satisfying max(g,h) < 6*gcd(g,h).

There are (at least) 6 solutions:

[ a, b] [ c, d] [ e, f] g/h

[ 1,47] [18,64] [24,62] 4/3

[ 9,32] [ 1,47] [19,43] 4

[12,31] [ 1,47] [23,41] 3

[19,43] [18,64] [46,48] 1/3

[23,24] [19,43] [23,41] 3/4

[23,41] [24,62] [46,48] 1/4

of which Jarek chose the last.

The other 5 solutions may be obtained from Jarek's,

by noting that

[18,64]/2 = [ 9,32]

[24,62]/2 = [12,31]

[46,48]/2 = [23,24]

and then permuting pairs of coprime integers.

Question: Are there any more solutions?

If not, then Jarek's construction is essentially unique.

Here is a test with b running up to 500:

{jarek(a,b)=local(s,t,C,G,c,d,g,h);

s=2*(a^2+b^2);t=sqrtint(s/2);C=[];G=[];

for(c=1,t,if(issquare(s-c^2,&d),g=abs((4*a*b)^2-(d^2-c^2)^2);

if(d>c&&g,G=concat(G,g);C=concat(C,[[c,d]]))));

for(i=2,length(G),h=G[i];for(j=1,i-1,g=G[j];

if(max(g,h)<6*gcd(g,h),print([[a,b],C[j],C[i],g/h]))))}

for(a=1,500,for(b=a+1,500,if(gcd(a,b)==1,jarek(a,b))))

[[1, 47], [18, 64], [24, 62], 4/3]

[[9, 32], [1, 47], [19, 43], 4]

[[12, 31], [1, 47], [23, 41], 3]

[[19, 43], [18, 64], [46, 48], 1/3]

[[23, 24], [19, 43], [23, 41], 3/4]

[[23, 41], [24, 62], [46, 48], 1/4]

Goodbye!

David