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.

 

I've promised a solution.  Here it  goes:

 

1. It is easy to show that only 2-digit numbers are eligible.
   
2. Assume the number  is  10*a+b
   Then  a^2+b^2+1 = 10*a+b
   a*(10-a)-1=b*(b-1)
   
3. From the above equation we can state:
   -If a   is a solution ,so  is  10-a (symmetry)
   -a is odd, since b*(b-1) is even
   
4. Now we solve an equation  b^2-b-k=0,
   where k=a*(10-a)-1:
   
   b=1/2*(1+sqrt(1+4*k)) ,   trying k=8,  20,  or  25
   (a=1 or 3 or 5)
   
5. k=20 triggers b=5, a=3 and 10-a=7
   
   
6. ANSWER: 35  and 75
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
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Edited on September 4, 2011, 4:00 am

Posted by Ady TZIDON on 2011-09-04 03:54:33