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We know whenever you square a given number $a$ The last digit of $a^2$ would be in the set :$$\{0,1,4,5,6,9\}$$ I didn’t quite understand the proof that was in my number theory book.

I will present it here: $$10 \equiv 0 \mod 10 \implies a=\sum_{i=0}^{n} a_i 10^i \equiv a_o \mod 10$$ $$\implies a^2\equiv a_o^2 \mod 10$$ But $a_o \in \{0,1,2,3...9\} \implies a_o^2 \mod 10 \in \{0,1,4,5,6,9\}$.

So the unit digit of $a^2$Is in the set $ \{0,1,4,5,6,9\} $ .

My problem is in the first line because if you call a function $$f(x)=\sum_{i=0}^{n} a_i x^i \implies f(0)=0$$

So why did the book named $a_o$ instead of $0$

PNT
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    No....$f(0)=c_0$, not $0$. – lulu Feb 18 '21 at 20:44
  • They are not evaluating a function at $0$ here, they are considering the number $a$ modulo $10$. Also I believe it should be $a=\sum_{i=0}^{n} a_i 10^i \equiv a_o \mod 10$ – Dylan Feb 18 '21 at 20:45
  • By the Polynomial Congruence Rule in the linked dupe we have

    $$ \bmod 10!:,\ \color{#c00}{10\equiv 0}\Rightarrow f(\color{#c00}{10})\equiv f(\color{#c00}0)\equiv a_0 = \text{constant coef of $f$}$$ Finally $,a\equiv a_0\Rightarrow a^2\equiv a_0^2,$ by the Congruence Power Rule.

     – Bill Dubuque Feb 18 '21 at 20:55
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    @Dylan - you’re right I’ve just checked the book it says $a_i$ not $c_i$ – PNT Feb 18 '21 at 21:52

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Note that when $f(x) = \sum_{j = 1}^n a_j x^j$, we have $f(0) = \mathbf{a_0}$, not $0$. Here $a_0$ represents the ones digit of the number $a$, where $a = a_n 10^n + \dots + a_1 10 + a_0$ in base $10$. Taking that last expression modulo $10$ kills all terms except the last, leaving $a \equiv a_0 \pmod{10}$.

Sam Freedman
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