Find the nearest biquadratic number
Python 3, 35 bytes



lambda n:int((n**.5-.75)**.5+.5)**4


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How it works

The value n at which the output switches from (k − 1)4 to k4 satisfies √(√n − 3/4) + 1/2 = k, or n = ((k − 1/2)2 + 3/4)2 = (k2 − k + 1)2 = ((k − 1)4 + k4 + 1)/2, which is exactly the first integer that’s closer to k4.

(Works for all n ≤ 4504699340341245 = (81924 + 81934 − 7)/2 > 252, after which floating-point roundoff starts to break it, even though it works mathematically for all n.)