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Sum Squared Divisors, Get Number (Posted on 2007-05-03) |
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Determine all possible positive whole numbers t such that t is equal to the sum of the squares of its four smallest positive divisors including 1.
For example, 1, 2, 4 and 5 are the four smallest positive divisors of 40. But, 12 + 22 + 42+ 52 = 46. So, t = 40 is NOT a valid solution.
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Submitted by K Sengupta
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Rating: 4.0000 (1 votes)
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Solution:
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t=130 is the only possible solution to the given problem.
EXPLANATION:
t cannot be odd. If so, then its four smallest divisors would be odd and hence the sum of their squares even.
So the two smallest divisors are 1 and 2.
If 4 divides t, then the four smallest divisors are either 1, 2, 4, 8 or 1, 2, 4, s ; where s may be 3 or 5 or more.
In the first case the sum of the squares is odd, but t is even, so it fails.
In the second case we have:
21 + s^2 = 4su, where u is the product of the other factors of t.
Now, any solution must divide 21, so s must be 3 or 7.
However, in that situation, 21 + s^2 is not divisible by 4, so this does not give any solutions.
So we can assume 2 divides n, but not 4. So if s is the smallest odd prime dividing t, then the three smallest divisors are 1, 2 and s.
If the next smallest is w, then three of the four smallest are odd and hence the sum of the squares is odd.
So the next smallest must be 2s.
Thus we have 5 + 5s^2 = 2su. So 5 divides t, and, accordingly, must be equal to 3 or 5.
s = 3 gives the sum of squares as 50, which is a contradiction as 50 is not divisible by 3.
s = 5 gives t = 130, which is thus the only possible solution to the given equation.
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