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¡Ñ26 = 130 = 1¡Ñ2¡Ñ5¡Ñ13, and the largest prime divisor is 13. _________________ ®]¤å¥ý ·q¤W
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