Positive Integers whose Square Root equals Sum of Digits

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Theorem

The following positive integers have a square root that equals the sum of their digits:

$0, 1, 81$

and there are no more.


Proof

\(\ds \sqrt 0\) \(=\) \(\ds 0\)
\(\ds \sqrt 1\) \(=\) \(\ds 1\)
\(\ds \sqrt {81}\) \(=\) \(\ds 9\)
\(\ds \) \(=\) \(\ds 8 + 1\)


By considering the square roots, we are looking for positive integers with a square for which its sum of digits is equal to the original number.

It is easy to verify the result up to $36$.


We prove that for $n > 36$, the sum of digits of $n^2$ cannot equal $n$.

Let $n$ be a $d$-digit number.

Then $n < 10^d$.

Thus $n^2 < 10^{2 d}$.

Therefore $n^2$ has at most $2 d$ digits.

Hence the sum of digits of $n^2$ is at most $18 d$.


If $n$ is a $2$-digit number, we have $n > 36 = 18 d$.

Now suppose $d \ge 3$.

Then:

\(\ds n\) \(\ge\) \(\ds 10^{d - 1}\)
\(\ds \) \(=\) \(\ds 10 \times 10^{d - 2}\)
\(\ds \) \(\ge\) \(\ds 10 \paren {1 + 9 \paren {d - 2} }\) Bernoulli's Inequality
\(\ds \) \(=\) \(\ds 90 d - 170\)
\(\ds \) \(\ge\) \(\ds 18 d + 216 - 170\) as $d \ge 3$
\(\ds \) \(>\) \(\ds 18 d\)

Therefore for $n > 36$, the sum of digits of $n^2$ cannot equal $n$.

$\blacksquare$


Sources

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