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Explanation:

a + b + c + d = 4m + 1

a2 + b2 = (a + b)2 – 2ab

a2 + b2 is minimum when 2ab is maximum.

The product of two numbers is maximum when the numbers are equal.

∴ a² + b² is minimum when a = b

Similarly, c² + d² is minimum when c = d

∴ a² + b² + c² + d² is minimum when a = b and c = d

∴ (a² + b² + c² + d²)min = 2(a² + c²)

But, a² + c² is minimum when a = c

∴ a² + b² + c² + d² is minimum when a = b = c = d

When a = b = c = d, a + b + c + d is a multiple of 4.

But, a + b + c + d = 4m + 1

So, one out of a, b, c, d must be one greater than the other three.

∴ a = b = c = m and d = m + 1

∴ a² + b² + c² + d² = m² + m² + m² + (m + 1)² = 4m² + 2m + 1

Hence, option (b).

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