Coefficients of the expansion of $(x+a)^2$ makes a perfect square?!

Question

I have no idea how this thought popped into my mind, but I noticed that the coefficients of $(x+1)^2$, when expanded, makes a perfect square. No, I am not talking about adding them (although that works too). Let me show you what I mean $$(x+1)^2=1x^2+2x+1$$ The coefficients are $1$, $2$, and $1$. Literally combining them into one big number makes $121$, which is a perfect square. This works for any terms $(x+a)^2$, provided that $a\in \mathbb N, \ a \le 9$. For example, $(x+3)^2=x^2+6x+9$, where the coefficients are $1$, $6$, $9$. When mashed together into one number, it makes $169$, which is $13^2$. I find this to be very interesting, and am wondering if there is a reason behind why this happens. Can someone please explain this to me? Thanks

Answer 1

This is because if you set $x=10$, you are getting $(10+a)^2 = 100+20a+a^2$, which is always a perfect square for integer values of $a$.

When $1\leq a\leq 4$ we set $x=10$ because by adding the components (ie: $100+40+4=144$ for $a=2$), we are literally just constructing a number from its digits.

But when $5 \leq a \leq 9$ we need to set $x=100$ because the middle term $20a$ is now a 3-digit number and again, by adding the numbers, we are pretty much just adding up the digits.

For example, $(100+6)^2 = 10000 + 1200 + 36 = 11236$, which is a perfect square of course.