Since (√2 + 1)^10 + (√2 − 1)^10 = u + f + v is an integer, and u is an integer, f + v must be an integer. Since 0 < f < 1 and 0 < (√2 − 1)^10 < 1 (as √2 − 1 ≈ 0.414 < 1), both f and v are between 0 and 1, so 0 < f + v < 2. The only integer in this range is 1. Therefore f + v = 1.
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