Some integers are one more than the sum of the squares of their digits in base 10.
Prove that the set of such integers is finite and list all of them.
Even though brute force is pretty quick I thought this was interesting because it shows the connection between the two solutions.
Any solution is clearly positive.
No solution can have 1 digit since no real is one less than its square.
As noted if we try 3 digits 9^2+9^2+9^2+1=244 so the first digit is 2 or less.
But then 2^2+9^2+9^2+1=167 so the first digit is 1 and the second is 6 or less.
But then 1^2+6^2+9^2+1=119 so the second digit is 1 or less.
But then 1^1+1^2+9^2=83 so there can be no 3 digit answer.
Any solution must have 2 digits.
Let the digits be A and B so we have
10A + B = A^2 + B^2 + 1
Solving for A by quadratic formula simplifies to:
A=5+/-sqrt(-B^2+B+24)
The only digit for B that makes the sqrt rational (a bit of brute force here) is B=5 which gives 2.
A=5+/-2 = 7 or 3
So the numbers are 75 and 35.
|
|
Posted by Jer
on 2015-12-08 11:25:09 |
Analytical solution
