<s>[INST] How find the minimum of the $abc$ let $a,b,c>2$ be prime numbers,such

$$a|b^5+1,b|c^5+1,c|a^5+1$$

Find the minimum of $abc$ 
My try:
I think use 
$$a^{p-1}=1(modp)$$
But I don't,Thank you for your help.
 [/INST] Note that $a, b, c$ are pairwise relatively prime and thus pairwise distinct. 
If $\min(a, b, c) \geq 7$, WLOG assume that $a=\max(a, b, c) \geq 13$, so $a>b \pm 1$. Now $b^5 \equiv -1 \pmod{a}, b^{10} \equiv 1 \pmod{a}, b \not \equiv \pm 1 \pmod{a}, b^2 \not \equiv 1 \pmod{a}$. Thus the order of $b \pmod{a}$ is $10$, so since by Fermat's little theorem $b^{a-1} \equiv 1 \pmod{a}$, we have $10 \mid a-1$. Thus $a \geq 31$. We get $abc \geq 7(11)(31)=2387$.
If $\min(a, b, c)=5$, WLOG assume that $a=\min(a, b, c)=5$. Then $c \mid 5^5+1=3126=6(521)$ so $c=521$ since $c>a=5$. Then $abc=5(521)b>5(521)(2)=5210$.
If $\min(a, b, c)=3$, WLOG assume that $a=\min(a, b, c)=3$. Then $c \mid 3^5+1=244=4(61)$ so $c=61$ since $c>2$. Now $5 \leq b, 3 \mid b^5+1$ and $b \mid 61^5+1$. We have $5 \nmid 61^5+1$ and $3 \nmid 7^5+1$ clearly so $b \geq 11$ and $abc \geq 3(11)(61)=2013$. Indeed $(a, b, c)=(3, 11, 61)$ works, so the minimum value of $abc$ is $2013$.
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