The integer cannot be odd. Else all its divisors would be odd and the sum of the four squares of its divisors even, yielding a contradiction. Hence the smallest two divisors must be 1 and 2.
The next smallest divisor must be 4 or a prime p. It cannot be 4.
Else the sum would include exactly two odd squares (to make it even) and would be a number divisible by 2 but not by 4, yielding a contradiction.
Thus, the smallest three divisors are 1, 2 and an odd prime p. Since the sum is even, the remaining divisor is 2p. Thus the number is equal to
1 + 4 + p^2 + 4p^2 = 5(1+p^2).
Since p does not divide 1 + p^2, it must divide (and so be equal to) 5.
The number is 5×26 = 130 = 1×2×5×13, and the largest prime divisor is 13.
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Re: 請教澳洲數學比賽的一道問題
2007-11-05 16:48The next smallest divisor must be 4 or a prime p. It cannot be 4.
Else the sum would include exactly two odd squares (to make it even) and would be a number divisible by 2 but not by 4, yielding a contradiction.
Thus, the smallest three divisors are 1, 2 and an odd prime p. Since the sum is even, the remaining divisor is 2p. Thus the number is equal to
1 + 4 + p^2 + 4p^2 = 5(1+p^2).
Since p does not divide 1 + p^2, it must divide (and so be equal to) 5.
The number is 5×26 = 130 = 1×2×5×13, and the largest prime divisor is 13.