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I have found a solution to 1b. We can express the two numbers which have a difference of 4 as x and x+4. We need to multiply the numbers and then add 4 on . Using our definition of the two numbers, we can write this process algebraically as x^2 + 4x + 4. What I notice is that this process gives the square of the mean of the two numbers, so x^2 + 4x + 4 = ((2x+4)/2)^2. This can be simplified to x^2 + 4x + 4 = x^2 + 4x + 4. This proves that if you multiply any two numbers which have a difference of four and if you add 4 on, the result is always the square of the mean of the two numbers.