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Step-by-step explanation:
1. let 2 consecutive odd integers = 2x+1 and 2x-1
(2x+1)^2 + (2x-1)^2 = 290
=> 4x^2 + 1 + 4x + 4x^2 + 1 - 4x = 290
=> 8x^2 = 299 - 2 = 288
=> x^2 = 36
=> x = ±6
=> x = 6 (as -6 is not positive integer so it's not a solution)
=> 2x+1 = 13
=> 2x-1 = 11
Hence there exists unique pair of consecutive positive odd integers I.e. 11 and 13 for the problem.
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