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Thus $5|(n^2 4^{n^2} +1)$

I tried to make it as

$n^2 4^{n^2} \equiv -1 \pmod 5$

$n^2 4^{n^2} \equiv 4 \pmod 5$

... suggest for how to do that

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  • $\begingroup$ You better learn how to type mathematics in this: it's very hard to understand what you wrote. $\endgroup$
    – DonAntonio
    Mar 18, 2020 at 16:10
  • $\begingroup$ Is that supposed to be $n^2\cdot 2^2\cdot n^2$, which is then $4n^4$ in the title? Your first line is different. $\endgroup$ Mar 18, 2020 at 16:12
  • $\begingroup$ It is n^2 * 2^(2n^2) +1 $\endgroup$ Mar 18, 2020 at 16:14
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    $\begingroup$ If you look at the patterns of final digit of $n^2$ and of $4^{n^2}$ then it becomes rather easy $\endgroup$
    – Henry
    Mar 18, 2020 at 16:44
  • $\begingroup$ When $4\equiv -1$ and if $n$ is even $4^{n^2}\equiv 1$ so you have $n^24^{n^2} + 1\equiv n^2 + 1$. If $n=2k$ then you have $n^2 + 1=4k^2 + 1\equiv 1-k^2\equiv pmod 5$. If $k\equiv 0,1,2,3,4$ then $1-k^2\equiv 1,0,-3,-3,0$. So $2$ ($2*(5k+1)=10k+2$ and $2*(5k+4)=10k + 8$) out of $5$ of the even numbers will qualify. So $20$. $\endgroup$
    – fleablood
    Mar 18, 2020 at 17:58

3 Answers 3

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We have that $n^24^{n^2}+1\equiv_5n^2(-1)^{n^2}+1$, also $n$ is even so that $$n^2+1\equiv_50\iff {n^2\equiv_5 -1}$$ This gives that $n\equiv_5 \pm2$. Since also $n\equiv_20$, we conclude that $$n\equiv_{10}\pm2$$

That is $$n\in S=\{2,8,12,18,22,28,32,38,42,48,52,58,62,68,72,78,82,88,92,98\}\\|S|=20$$


I've computed some values of $n^22^{2n^2}+1\text{ (mod }5)$ in the table below, using WolframAlpha. $$\begin{array}{|l|l|} \hline n& n^22^{2n^2}+1\text{ (mod } 5) \\ \hline 1& 0\\ \hline \color{green}{2}& \color{green}{0}\\ \hline 3& 2\\ \hline 4& 2\\ \hline 5& 1\\ \hline 6& 2\\ \hline 7& 2\\ \hline \color{green}{8}& \color{green}{0}\\ \hline 9& 0\\ \hline 10& 1\\ \hline 11& 0\\ \hline \color{green}{12}& \color{green}{0}\\ \hline 13& 2\\ \hline 14& 2\\ \hline 15& 1\\ \hline \end{array}$$

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If $n$ is an even number, then $n^2$ is divisible by $4$. Now, let $m=n^2$, so that $m \cdot 2^{2m}$ must then also be divisble by $4$. If $m \cdot 2^{2m}+1$ is divisible by $5$, then $m \cdot 2^{2m}$ must end with a $4$ (it could not end with a $9$ because it is divisible by $4$).

Unless $m=0$, $2^{2m}=4^m$ must always end with a $6$ since $m$ is even. Hence, it suffices to check the last digit of $m$.

Squares never end with the digits $2$ or $8$ (actually, $3$ and $7$ are also impossible last digits of squares, but they are odd).

If $m$ ends with a $0$ (so that $n$ also ends with a $0$), then $m \cdot 2^{2m}$ must once again end with a $0$.

If $m$ ends with a $4$ (so that $n$ ends with either a $2$ or an $8$), then $m \cdot 2^{2m}$ must also end with a $4$.

Finally, if $m$ ends with a $6$ (so that $n$ ends with either a $4$ or a $6$), then $m \cdot 2^{2m}$ must also end with a $6$.

The answer is therefore the number of even numbers under a hundred ending with either a $2$ or an $8$, which is $\mathbf{20}$.

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  • $\begingroup$ $n$ doesn't have to be square. So $n$ could end with $8$. $\endgroup$
    – fleablood
    Mar 18, 2020 at 17:58
  • $\begingroup$ @fleablood Yes, but $m$ is a square, so it cannot end with a $2$ or an $8$. $\endgroup$ Mar 18, 2020 at 18:03
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Muck it out.

$n^2 2^{2n^2}+1 = n^24^{n^2}+1 \equiv n^2(-1)^{n^2} + 1 \equiv \begin{cases}-n^2 +1 &\text{if }n\text{ is odd}\\n^2 +1&\text{if }n\text{ is even}\end{cases}\pmod 5$.

If $n$ is even then let $n=2k$ and $n^2=4k^2 \equiv -k^2$and let $k\equiv 0,\pm 1, \pm 2$ then $n^2 + 1\equiv -k^2 + 1\equiv 1,0,2\pmod 5$.

So $k\equiv \pm 1 \pmod 5$ so $k= 5m \pm 1$ and $n = 10m \pm 2$ so there are $20$ such even numbers $2,8,12,18,...etc...$.

If $n$ is odd then let $n = 2k+1$ and $n^2 = 4k^2 + 4k +1\equiv -k^2 -k+1$ and $-n^2+1\equiv k^2 + k\pmod 5$. If $k\equiv 0, 1,2,-2,-1$ then $k^2 + k\equiv 0,2,1,2,0$ so $k \equiv 0,-1\pmod 5$ for $k = 5m +(0,-1)$ and $n = 10m +1 (-2)=10m\pm 1$. So odd numbers $n = 1,9,11,19$ etc.

I know you specified that $n$ is even but... we might as well have figure them all out.

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