淺見提供，有錯還望不吝指教
若\(\displaystyle x=7p+1 \ or \ 7p+6 \Rightarrow x^4\equiv 1\ (mod\ 7)\)
若\(\displaystyle x=7p+2 \ or \ 7p+5 \Rightarrow x^4\equiv 2\ (mod\ 7)\)
若\(\displaystyle x=7p+3 \ or \ 7p+4 \Rightarrow x^4\equiv 4\ (mod\ 7)\)
其中\(\displaystyle p\in \mathbb{N} \cup \{0 \}\)
並且
若\(\displaystyle x=3k+1 \Rightarrow 4^x\equiv 4 \ (mod\ 7)\)
若\(\displaystyle x=3k+2 \Rightarrow 4^x\equiv 2 \ (mod\ 7)\)
若\(\displaystyle x=3k \Rightarrow 4^x\equiv 1 \ (mod\ 7)\)
其中\(\displaystyle k \in \mathbb{N} \cup \{0 \}\)
接下就是分類
\(\displaystyle x=3k=7p+1 \Rightarrow x=15+21N\)
\(\displaystyle x=3k=7p+6 \Rightarrow x=6+21N\)
\(\displaystyle x=3k+1=7p+3 \Rightarrow x=10+21N\)
\(\displaystyle x=3k+1=7p+4 \Rightarrow x=4+21N\)
\(\displaystyle x=3k+2=7p+2 \Rightarrow x=2+21N\)
\(\displaystyle x=3k+2=7p+5 \Rightarrow x=5+21N\)，以上個數在範圍之下皆為10個，故共有60個數滿足題意
PS. \(x\)為7的倍數必無法滿足題意，故不考慮
