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Squaring Up The Digits II (Posted on 2011-09-03) Difficulty: 3 of 5
Determine all possible values of a base ten positive integer N such that N is precisely one more than the sum of the squares of its digits.

No Solution Yet Submitted by K Sengupta    
Rating: 4.3333 (3 votes)

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Possible soution | Comment 4 of 7 |
a^2+b^2+1=10a+b
4(a-5)^2+(2b-1)^2 = 97
say, (2x)^2+y^2=97
x=+/-2, y=+/-9
a-5=2,a-5=-2
a={7,3}
2b-1=9, (2b-1=-9)
b={5}
N={35,75}
  Posted by broll on 2011-09-04 01:29:59
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