<s>[INST] Prove that $7^{100}+3^{10}=8^{100}$ or $7^{100}+3^{10}<8^{100}$ 
Prove that $7^{100}+3^{10}=8^{100}$ or $7^{100}+3^{10}<8^{100}$

I tried using some theorems of divisibility, to show that one divides the other, and the other also divides the first, but could not.
 [/INST] If you list $7^x$ and $8^x$ you will notice that when $x$ is divisible by $4$, $7^x$ ends in $1$ and $8^x$ ends in $6$.  $$7^0=8^0=1$$ $$7^{4+j}=7^j*7^4$$ $$7^4 = 2401$$ $2401$ ends in $1$ as does $7^0$ and any number that ends in $1$ times any number than ends in 1 results in a number that also ends in $1$. This works for $8$ as $8^4$ ends in $6$ and $6$ has the same property. 
In order for $7^x+y=8^x$ then $y$ must end in $5$ which is impossible for $3^z$ as any number that ends in $5$ is divisible by $5$.   
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