For many integers $n$, the digit sum of $(4n^2+1)^2$ is $19$. Is there something special about $19$?

Question

^{I will denote by $\mathbb{N}$ the set of all natural numbers $>$ $0$; id est, the set $\{1,2,3,\ldots\}$. Also, "$u\mid v$" is read as "$u$ divides $v$".}

There seem to be many numbers $n\in\mathbb{N}$ such that $(4n^2+1)^2$ has a digit sum of $19$. $\; (1)$

List of such numbers $n$ I have found: $$\begin{align}&(4\times \phantom{0}2^2+1)^2=\phantom{00000}289\to 2+8+9=19.\tag{$n=2$} \\ &(4\times \phantom{0}3^2+1)^2=\phantom{0000}1369\to 1+3+6+9=19.\tag{$n=3$} \\ &(4\times 11^2+1)^2=\phantom{00}235225\to 2+3+5+2+2+5=19.\tag{$n=11$} \\ &(4\times 15^2+1)^2=\phantom{00}811801\to 8+1+1+8+0+1=19.\tag{$n=15$} \\ &(4\times 16^2+1)^2=\phantom{0}1050625\to 1+0+5+0+6+2+5=19.\tag{$n=16$} \\ &(4\times 20^2+1)^2=\phantom{0}2563201\to 2+5+6+3+2+0+1=19.\tag{$n=20$} \\ &(4\times 21^2+1)^2=\phantom{0}3115225\to 3+1+1+5+2+2+5=19.\tag{$n=21$} \\ &(4\times 24^2+1)^2=\phantom{0}5313025\to 5+3+1+3+0+2+5=19.\tag{$n=24$} \\ &(4\times 25^2+1)^2=\phantom{0}6255001\to 6+2+5+5+0+0+1=19.\tag{$n=25$} \\ &(4\times 29^2+1)^2=11323225\to 1+1+3+2+3+2+2+5=19.\tag{$n=29$}\end{align}$$ Then we have the special case $n=31$ where the digit sum adds up to itself!

But my question is, is this just a coincidence? Or is there some explanation? I know that $4n^2=(2n)^2$, though this does not help.

If we consider $(2n^2+1)^2$ for example, then the digit sum of that expression across many natural numbers $n$ are multiples of $9$. But that is understandable $-$ any number with a digit sum of a multiple of $9$ must also be divisible by $9$ (and the same goes with $3$).

And $3\mid 2n^2+1\Leftrightarrow 3\mid 2n+1\Leftrightarrow 3\mid n-1$. (If $3\mid 2n^2+1$ then $3\mid 2n^2-2$ and now factorise, etc).

So I can find digit sums of multiples of $9$ by just letting $n\equiv 1\pmod 3$. But, what about for $(1)$? How do I find out what $n$ to choose with some division?

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Thank you in advance.

Answer 1

This seems to be a "small number" effect. None of $[25\,001, 150\,000)$ have digit sum $19$. In fact, of $n \in [101, 10^6]$, only $106$, $150$, $200$, $250$, $1\,500$, $2\,000$, $2\,500$, $15\,000$, $20\,000$, $25\,000$, $150\,000$, $200\,000$, and $250\,000$ give digit sums of $19$. This is a pretty compelling pattern, and it continues without omissions or additions up to $10^9$ (easy enough to check in software for each $n$ up to $10^9$), but I haven't tried to prove anything about it.

(I'm not surprised. For the digit sum of a long number to be $19$, most of its digits must be zeroes. The set of $n$ that will produce all these zeroes gets thinner and thinner as we go. That "$15$ and some zeroes", "$20$ and some zeroes", and "$25$ and some zeroes" will give digit sums of $19$ is already visible in your table.)

Answer 2

Well, if $n \equiv 0,\pm 1,\pm 2,\pm 3,\pm4\mod 9$ then $n^2\equiv 0,1,4,0,-2$ and $4n^2+1\equiv 1,5,-1,1,2$ and $(4n^2 + 1)^2 \equiv 1,-2,1,1,4\mod 9$ so a lot of numbers (5 out of 9) will have numbers whose digits add to $1$ more than a multiple of $9$.

As the first several numbers are kind of small but not that small it's not surprising that their digits are more than $10$ but less than $28$.

If we check the numbers below $31$ that you didn't check that are $\equiv 0,2,3,6,7 \mod 9$ are: $6,7,12, 18, 27,30$. Which go to $10$ (too small), $28$,$28$, $28$, $46, 28$ .

So $\frac 59$ of the numbers less than $31$ are potential. Of those $21$, $15$ are in the sweet range and $6$ are not. That's .... plausible.