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I'm almost certain this is an inductive proof. I've expressed it as for all $n\in2x$ and $\notin10y$ , $n^4\equiv 6\mod10 \forall x,y\in\Bbb Z^+$. The base case is straightforward, as $2^4=16\equiv6\mod10$. Just struggling on proving for when $n=k+2$. Binomial expansion gives $k^4+8k^3+24k^2+32k+16$, which implies $k^4+8k^3+24k^2+32k\equiv0\mod10$, which using the assumption $k^4\equiv6\mod10$ gives $8k^3+24k^2+32k\equiv4\mod10$.

Any help in going forward or which direction to take would be excellent.

Ted Shifrin
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Daniel Jago
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  • Your induction is going to have trouble when $n\equiv 8$ and $n^4 \equiv 6$ but $(n+2)^4 \equiv 0$. – fleablood Mar 31 '21 at 05:26
  • Since $f(n) = n^4-6$ satisfies $f(n+10)\equiv f(n)$ you can use (complete, strong) induction as in the dupes to reduce it to the base case(s) of nonzero evens $< 10\ \ $ – Bill Dubuque Mar 31 '21 at 15:38

3 Answers3

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No need for that much trouble. This is a proof by enumeration. Proving for $n=2,4,6,8$ is enough.

trisct
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    In fact, $ \ 2 \ $ and $ \ 6 \ $ are enough, since $ \ 2^2 \ $ and $ \ 2^3 \ $ will end as $ \ 16^2 \ $ and $ \ 16^3 \ $ . –  Mar 31 '21 at 02:46
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Your induction approach is close.

Instead of the single base case of $2$, you can use the manual base cases of each element in $\{2,4,6,8\}.$ As other responses have suggested, you can even argue (for example) that $8^4 \equiv (-2)^4 \equiv 2^4 \pmod{10}.$

Then, having established the base cases, you can use induction by proving that if the assertion is true for $N$, then it must also be true for $(N+10)$.

This can be demonstrated, simply by examining the binomial expansion of $(N + 10)^4$ and comparing it with $N^4 \pmod{10}.$

This demonstration coupled with your manual examination of the base cases of $\{2,4,6,8\}$ will be conclusive.

user2661923
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Your induction is going to have trouble when $n\equiv 8$ and $n^4 \equiv 6$ but $(n+2)^4 \equiv 0$.

You can do it individually. $2^4 \equiv 16\equiv 6$ and $4^4\equiv 256 \equiv 6$ and $6^4\equiv 1296\equiv 6$ and $8\equiv 4096\equiv 6$.

If you don't want to do the four cases individually do this

$(-d)^4 = d^4$ and $8\equiv -2\pmod {10}$ and $6\equiv -4$. And just do $2$ and $4$.

Or if you want to really cut off you nose to spite your face:

Note $n= 10a \pm (5b \pm c)$ where $b=0,1$ and $c=2,1$ (respectively)

Square it and you get something equiv $25b^2 \pm 10bc + c^2\pmod {10}$ but $25\equiv 5$ and $b^2 \equiv 0^2, 1^2\equiv b$. And $c^2 \equiv (5-1), 1$. So if $b \equiv 0,1$ the $c^2 \equiv 5-1, 1$ respectively.

So $n^2 \equiv 5b + c^2 \equiv 5 \pm 1$.

And so $n^4$ equiv $(5\pm 1)^2 = 25 \pm 10 + 1\equiv 6\pmod{10}$

fleablood
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