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Find the lowest n (Posted on 2015-06-22)

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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A few other n

| Comment 2 of 4 |

20 = (1+4)*1*4

1/20 = 1/5^2 + 1/10^2

Some other n that work include

90 = (1+9)*1*9

1/90 = 1/10^2 + 1/30^2

272 = (1+16)*1*16

1/272 = 1/17^2 + 1/68^2

(1+a^2)*a^2

1/(a^2 +a^4) = 1/(1+a^2)^2 + 1/(a+a^3)^2

Posted by Steve Herman on 2015-06-23 06:47:48

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