Digit sum equals 44 of a squared number

Question

Is there a whole number $x$ such that the sum of the digits of $x^2$ equals 44?

I would like someone to tell me if my thoughts are correct.

The remainder of a number a divided with 9 is the same as the remainder of the sum of digits of a divided with 9.

According to that we have:

$\frac{x^2}{9}=d+r$

and

$\frac{44}{9}=d_1+r$

If there existed such a whole number $x$ then $d_1$ would be a whole number, which it is not. So I think the answer should be no.

Answer 1

Hint: If the sum of the digits of $x^2$ is $44$, then $x^2$ has remainder $2$ on division by $3$. This is impossible.

Remark: Your proposed argument is not correct. Please note that one could use exactly the same argument to "show" that the sum of the digits of $x^2$ cannot be $7$. But it can be: the sum of the digits of $16$ is $7$.

Answer 2

You would write $\dfrac{x^2}{9} = d + \dfrac {44}9$, which gives you $x^2 = 9d + 44$. Since $\dfrac{44}9 = 4 + \dfrac 89$, you could combine the two and get $x^2 = 9(d+4) + 8$. None of these seem to imply any contradictions.

The sum of the digits of $0$ through $8$ squared are

 0 → 0
 1 → 1
 2 → 4
 3 → 9
 4 → 16 → 7
 5 → 25 → 7
 6 → 36 → 9
 7 → 49 → 13 → 4
 8 → 64 → 10 → 1

Every positive integer is of the form $9a + b$ where $0 \le b \le 8$.

It follows that, if you keep summing the digits of $x^2$ until you get just one digit, that digit is going to have to be $0, 1, 4, 9, $ or $7$. the sum will never be $44$, whose digits sum to $8$.