<s>[INST] Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$ Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$
I tried to induct on n:
For $n = 0$ we have $3+5 = 8$ and $8 \equiv 1 \pmod{7^{n+1}}$.
Suppose it is true for $n = k$: 
$$3^{7^k}+5^{7^k}\equiv 1 \pmod{7^{k+1}}$$
so $3^{7^k}+5^{7^k}=7^{k+1}*q_1+1$
For $n = k+1$:
$$3^{7^{k+1}}+5^{7^{k+1}}\equiv r \pmod{7^{k+2}}$$
so $3^{7^{k+1}}+5^{7^{k+1}}=7^{k+2}*q_2+r$
Now if we subtract them we get that:
$$3^{7^{k+1}}+5^{7^{k+1}}-3^{7^k}-5^{7^k}=7^{k+1}(7q_2-q1)+r-1$$
From Euler's theorem we know that $a^{\phi(7^{k+1})}\equiv 1 \pmod{7^{k+1}}$ 
and $\phi(7^{k+1}) = 6\cdot7^{k}$ so $3^{6\cdot7^{k}}\equiv 1 \pmod{7^{k+1}}$ and $5^{6 \cdot 7^{k}}\equiv 1 \pmod{7^{k+1}}$.
So $3^{7^{k+1}}+5^{7^{k+1}}-3^{7^k}-5^{7^k}\equiv3^{7^k}\cdot3^{6\cdot{7^{k}}}+5^{7^k}\cdot5^{6\cdot{7^{k}}}-3^{7^k}-5^{7^k}\equiv 0 \pmod{7^{k+1}}$
Finally I get that $r\equiv1\pmod{7^{k+1}}$.
Here I got stuck. I don't know how to show that $r=1$.
 [/INST] Quick solution using Lifting the Exponent Lemma:
Note that $3^{7^n} + 5^{7^n} - 1 \equiv -4^{7^n} - 2^{7^n} - 1 \equiv -(4^{7^n} + 2^{7^n} + 1) \equiv -\dfrac{8^n-1}{2^n-1} \pmod{7^{n+1}}$.
Now, by LTE, we have $\nu_7(8^{7^n} - 1) = \nu_7(8-1) + \nu_7(7^n) = n+1$, and $2^{7^n} \equiv 2^{7^n \pmod{6}} \equiv 2^{1^n} \equiv 2 \pmod{7}$, so $7 \nmid 2^{7^n} - 1$. Thus, we have
$$\frac{8^{7^n}-1}{2^{7^n} - 1} \equiv 0 \pmod{7^{n+1}}$$
so we're done.
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