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I am pretty sure you can figure this out on your own. Raise different numbers to the square, until you get a 4-digit result. Similary, calculate the cube of different numbers, until you get a 4-digit number. If you want the SAME number to be both a perfect square and a perfect cube, then it must be a power of 6. In that case, just experiment raising different numbers to the sixth power, until you get a 4-digit number.
No.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theorem
Almost perfect numbers refer to numbers whereσ(x) = 2x - 1, where σ is the sum of divisors function. Any number in the form 2n is almost perfect becauseσ(2n) = 1 + 2 + 4 + ... + 2n = 2n+1-1 = 2(2n) - 1.It is unknown whether any other almost perfect numbers exist.