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Find the lowest n (Posted on 2015-06-22) Difficulty: 3 of 5
Are there distinct positive integers a,b,n for which:
1/n = 1/a^2+1/b^2,
?

If so, find the triplet with lowest n.

No Solution Yet Submitted by Ady TZIDON    
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Solution Analytical solution Comment 4 of 4 |
We can state that b>a

It follows that either:
(1) b=c*a (c is positive integer)  or
(2) b=(p/q)*a (p,q positive integers)

In case (1) we can reduce the expression: n = a^2 * c^2 / (c^2 +1)
This means that a^2 = 0 mod (c^2+1)

Clearly c=1 doesn't work but with c=2 we find the previously presented solution (a=5, b=10, n=20)

No need to look beyond this because raising c will only grow a,b & n, thus it doesn't help in finding any smaller n

In case (2) we can reduce the expression to be : n = a^2 * (p/q)^2 / (1+(p/q)^2)
This also means that a^2 = 0 mod (1+(p/q)^2), since n has to be integer and (p/q)^2 not dividing (1+(p/q)^2)

From this we can see that (p/q) has to be integer, which means that the solution found in case (1) results into the smallest possible n!

  Posted by atheron on 2015-06-23 07:52:28
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