Are there squares S = n
2 such that when you add S and its digits the result is also a square?
For example, if S were 25 then 25+2+5 would also have to be a square.
If so, give an example.
If not, prove it.
Consider the first few values of n:
n S S+sod(S) (n+1)^2
1 1 2 4
2 4 8 9
3 9 18 16
4 16 23 25
5 25 32 36
Except for n=3, adding sod(S) to S will never be enough to get to the next higher square.
If we can prove that sod(n^2) < 2n + 1 for n greater than some value then we are done.
If n has 'd' digits then n^2 has either 2d or 2d-1 digits; take the larger: 2d.
The largest possible sod(n^2) is then 18d.
The smallest possible n with d digits is 10^(d-1) (though n^2 would then only have 2d-1 digits, but I'm being generous).
if n = 10^(d-1) then 2n+1 = 2*10^(d-1) + 1
If d > 2 then the inequality holds true:
18d < 2*10^(d-1) + 1
d 18d 2*10^(d-1) + 1
1 18 3
2 36 21
3 54 201
4 72 2001
5 90 20001
A computer check shows it is not true for any value under 100 and the argument above shows it is not true for any n with 3 or more digits.
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Posted by Larry
on 2024-03-12 12:27:43 |
Solution
