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So I have the simple polynomial $x^2+1$. If I plug in ANY number that has a $3$ or a $7$ in the ones place $x^2+1$ is divisible by $10$. Why? Is there a reason for this?

So numbers like $3,7,13,17,23,27,\ldots$ when plugged into $x^2+1$ is divisible by 10. Why? Is there a reason for this?

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    $\begingroup$ $x$ ends with a $3$ means $x^2$ ends with a $9$. $x$ ends with a $7$ means $x^2$ ends with a $9$. Now add 1. They both end with a $0$ $\endgroup$ – George Feb 10 '16 at 18:48
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\begin{align*} (10k + 3)^2 + 1 &= 100k^2 + 60 k + 9 + 1 = 10(10k^2 + 6k + 1) \end{align*}

and similarly for $7$.

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  • $\begingroup$ Alternatively, one could do $x^2+1\equiv 3^2+1\equiv 0\mod 10$. Nice solution though, +1 $\endgroup$ – vrugtehagel Feb 10 '16 at 18:50
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    $\begingroup$ We can write this, incidentally, as $(10k \pm 3)^2+1 = 100k^2 \pm 60k + 9 + 1 = 10(10k^2\pm 6k+1)$, and take care of both cases. It also shows how they're related. $\endgroup$ – Brian Tung Feb 10 '16 at 18:52
  • $\begingroup$ @BrianTung Yes, a nice observation; and for even more generality this is related to the fact that $k^2$ and $(n - k)^2$ have the same residue mod $n$. $\endgroup$ – user296602 Feb 10 '16 at 18:53
  • $\begingroup$ I got it. Thanks :) . $\endgroup$ – Future Math person Feb 10 '16 at 18:56

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