Solution: 

Same number four times, 2 × 2 × 2 × 2 = 2² + 2² + 2² + 2² = 16. 

Same number three times, 6 × 2 × 2 × 2 = 6² + 2² + 2² + 2² = 48. 

Same number two times, 22 × 6 × 2 × 2 = 22² + 6² + 2² + 2² = 528. 

Four different numbers, 262 × 22 × 6 × 2 = 262² + 22² + 6² + 2² = 69,168. 

The solution (2, 2, 2, 2) is rather easy to find. It can be found using a guess‑and‑check strategy, or you can solve it algebraically with the equation x⁴ = 4x². That equation simplifies to x² = 4, so x = 2. (It turns out that x = ‑2 is also a solution to that equation, but the problem asks for positive integers.)

In general, if (a, b, c, d) is a solution to the equation, then (bcd – a, b, c, d) is also a solution. That is, the value of a can be replaced by the value of bcd – a. It may not be obvious why this is true, but the following algebra shows that if a is replaced by bcd – a, then the equation still holds true:  

Consequently, if (a, b, c, d) = (2, 2, 2, 2), then bcd – a = 2 × 2 × 2 – 2 = 6 can replace a to yield the solution (6, 2, 2, 2). This solution uses one number three times.

The order of the numbers is unimportant, so (6, 2, 2, 2) = (2, 6, 2, 2) = (2, 2, 6, 2) = (2, 2, 2, 6). Taking the second of these as (a, b, c, d) = (2, 6, 2, 2), then bcd – a = 6 × 2 × 2 – 2 = 22 can replace a to yield the solution (22, 6, 2, 2). This solution uses one number twice.

By similar reasoning, 22 × 6 × 2 – 2 = 262 can replace a in (2, 22, 6, 2) to give the solution (262, 22, 6, 2), which uses four different numbers.