Each of a, b, c, d and e is a positive integer satisfying this system of equations:

a^-2 + b^-2 + c^-2 = d^-2, and:

a+b+c = e²

Determine the smallest value of e. What are the next two smallest values of e?

Restating the problem: 1/(a^2) +1/(b^2) + 1/(c^2)= 1/(d^2)   
Equivalently: a^2d^2(b^2+c^2)=b^2c^2(a^2-d^2)   
   
One set of solutions is a=3n,b=3n,c=6n,d=2n;    
a+b+c=12n=e^2: e={6,12,18…}   
   
Another is a=7m,b=14m,c=21m, d=6m   
(7m)^2d^2((14m)^2+(21m)^2)=(14m)^2(21m)^2((7m)^2-d^2)   
Hence 31213d^2m^4 = 86436m^4 (7m-d) (7m+d)   
a+b+c=(42n)^2; e={42,84,126,168..}   
   
There is also, e.g.   
(11n)^2d^2((22n)^2+(33n)^2)=(22n)^2(33n)^2((11n)^2-d^2)   
456989533d^2m^4 =1265509476m^4 (77m-d) (77m+d)   
d=66m, n=7m  etc., with rather large e.   
   
So the 3 smallest solutions are e= {6,12,18}   

Posted by broll on 2011-05-27 12:46:03