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Diophantine quickie (Posted on 2011-05-27) Difficulty: 2 of 5
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?

No Solution Yet Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Possible solution | Comment 1 of 3
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   
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.   
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
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