Show that $61!+1$ is divisible by $71$

Question

I'm working on the following Wilson's theorem exercise.

Show that $61!+1$ is divisible by $71.$

I know that expressed as a congruence it would be $61! \equiv -1 \pmod {71}$

Based on Wilson's theorem $(p-1)! \equiv -1 \pmod p$, I have

$$70! \equiv -1 \pmod {71}.$$

What I'm doing now is

$$70*69*68*...*62*61! \equiv -1 \pmod {71}.$$

But I don't know how to cancel those extra numbers I have before the 61; any hint or help will be greatly appreciated.

Answer 1

Notice: $62*..... *70 \equiv (-9)*(-8)*.... *(-1)\equiv$

$(9*8)*(6*4*3)*(7*5*2)*(-1)^9 \equiv -(72)^2(70)\equiv -1^2*(-1) \equiv 1\pmod {71}$.

So $70! = 61!*(62*....*70) \equiv 61! \pmod {71}$.

But $71$ is prime so by Wilson's Theorem $70! \equiv - 1\pmod {71}$.

So $61! \equiv -1 \pmod{71}$ and so $71|61! + 1$.

Answer 2

$$-1 \equiv 70\times69\times68\times...\times62\times61! \equiv -1\times-2\times\cdots\times-9\times61!\equiv-9!\times61!\pmod {71}$$

Now, we need to show $-9!\equiv1\pmod{71}$, which is $362881\equiv0\pmod{71}$ which is true.

Answer 3

Notice that (the following are all $\mod 71$): $$70 \equiv -1 $$ $$69 \equiv -2 $$ $$...$$ $$62 \equiv -9$$ $$70! \equiv (-1)(-2)...(-9)61! \equiv (-1)*61!$$

Answer 4

I don't know whether there is some really clever way, but here a brute force approach is fast enough. $$ 70\cdot69\cdots62\equiv (-1)\cdot(-2)\cdots(-9) $$ We have $$(-8)\cdot(-9)\equiv 1\\(-2)\cdot (-5)\cdot(-7)\equiv 1\\(-3)\cdot(-4)\cdot(-6)\equiv-1$$which means that $70!\equiv61!$