Squares ending in 6: Why do they have this in common?

Question

If you remove the last digit of a square number (for example $15^2=(22)5,13^2=(16)9$), why is it that when that number is odd ($16^2=(25)6$) then the final digit is always 6?

I'm aware that due to this there's only a limited number of final digits. But why does only the squares of 4 and 6 have an odd "prefix"

I assume it has something to do with the fact that the single digit squares only 4 and 6 have an odd prefix

1 01
2 04
3 09
4 16 -> odd
5 25
6 36 -> odd
7 49
8 64
9 81

But I don't see the connection.

Just as the last digit has a pattern, so do the last two. Cf. here. — Parcly Taxel, Sep 19 '22 at 07:30
"Why" is always a difficult question to nail down in mathematics. It is easy to prove that this is the case (you have basically a full proof listed there already), but coming up with a fundamental reason beyond "Just because" is gonna be hard. You can prove that there is a pattern, but explaining why the pattern has to be what it is, other than just saying "it is what it is and can't be anything else" is, I think, basically impossible. — Arthur, Sep 19 '22 at 07:32
Does this answer your question? Show that the last two decimal digits of a perfect square must be one of the following pairs. — Anne Bauval, Sep 19 '22 at 07:33
One pretty much has to use brute force, but a clever trick would be to narrow down the amount of cases required to check $\pmod {100}$. — Hersh, Sep 19 '22 at 08:30
@Hersh They have basically checked all the cases they need to check already right there in the question post. Remember, we only care about the parity of the ten's digit. — Arthur, Sep 19 '22 at 08:39

score 1 · Answer 1 · answered Sep 19 '22 at 08:56

For any real number $~r~$, let $\lfloor r\rfloor$ denote the largest integer $~\leq r$.

For $a \in \Bbb{Z^+}, b \in \{0,1,2,\cdots,9\}$, I claim that

$$\left\lfloor\frac{(10a + b)^2}{10}\right\rfloor$$

is odd if and only if

$$\left\lfloor \frac{b^2}{10}\right\rfloor$$

is odd.

Proof

$$\left\lfloor\frac{(10a + b)^2}{10}\right\rfloor $$

$$= \left\lfloor\frac{(100a^2 + 20ab + b^2}{10}\right\rfloor \tag1 $$

$$ = 10a^2 + 2ab + \left\lfloor \frac{b^2}{10}\right\rfloor. \tag2 $$

The expressions in (1) and (2) above are equivalent because of the following principle:

if $P \in \Bbb{Z}$ and $r \in \Bbb{R}$, then

$$\left\lfloor P + r\right\rfloor = P + \lfloor r \rfloor.$$

Further, in (2) above, each of the first two terms is an even number. Therefore, the overall sum will be an even number if and only if the 3rd term is even.

With the claim established, the remainder of the answer to the posted question is supplied by the analysis already given by the OP (i.e. original poster).

That is, the OP has shown that for $b \in \{0,1,\cdots,9\}$, the only instances when

$$\left\lfloor \frac{b^2}{10}\right\rfloor$$

is odd are when $b \in \{4,6\}.$ Further, these are (also) the only instances where the rightmost (i.e. one's) digit of $b^2$ is $(6)$.

Squares ending in 6: Why do they have this in common?

1 Answers1