Knowing that $a,b$ are prime integers and $a+b\equiv 1 \pmod{7^{n+1}}$ show that $a^7+b^7\equiv 1 \pmod{7^{n+2}}$
I used $a^7+b^7=(a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)$ and tried to show that $(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)\equiv 1 \pmod 7$ but after some trial and error I figured that it might not help as much as I thought. I also tried to show that $ a^7+b^7-1\equiv 0 \pmod{7^{n+2}}$, but I got stuck.
How should I solve this?
It's not true. Take $a=7, b=43$. Then $a+b\equiv 1\bmod 49$, but $a^7+b^7\equiv 295\bmod 343$.
In fact, I would think that the primality or otherwise of $a$ and $b$ is irrelevant, given that we are only interested in their values mod $7^{n+1}$; but perhaps if you require that they be co-prime to $7$, your result might hold.
Edited to add: No, that doesn't work either. Take $a=53,b=193$ (both prime, and both co-prime to $7$). Then $a+b\equiv 1\bmod 49$, but $a^7+b^7\equiv 134\bmod 343$. So it seems your claim is simply false.
