The beauty of 71

Consider the binomial expansion of $(a+1)^{71}$ and $(a-1)^{71}$ which gives these two expressions below respectively:

$a^{71} + { 71 \choose 1} a^{70} + { 71 \choose 2} a^{69} + \ldots + { 71 \choose 70} a^{1} + { 71 \choose 71} a^0$

$a^{71} - { 71 \choose 1} a^{70} + { 71 \choose 2} a^{69} - \ldots + { 71 \choose 70} a^{1} - { 71 \choose 71} a^0$

Take their difference and we are left with

$2 \bigg [ { 71 \choose 1} a^{70} + { 71 \choose 3} a^{68} + + { 71 \choose 5} a^{66} + \ldots + { 71 \choose 69} a^{2} + { 71 \choose 71} a^{0} \bigg ]$

In this case, $a = \sqrt{71}$

$2 \bigg [ { 71 \choose 1} 71^{35} + { 71 \choose 3} 71^{34} + { 71 \choose 5} 71^{33} + \ldots + { 71 \choose 69} 71^{1} + { 71 \choose 71} \bigg ]$

Consider modulo $10$ on the powers (for any natural number $n$ ): $(71)^n \equiv (71 \bmod 10)^n \equiv 1$ gives

$2 \bigg [ { 71 \choose 1} + { 71 \choose 3} + { 71 \choose 5} + \ldots + { 71 \choose 69} + { 71 \choose 71} \bigg ]$

We are only adding the odd terms of the $71^\text{st}$ row of the Pascal Triangle

$2 \cdot \left ( \frac {1}{2} \cdot 2^{71} \right ) = 2^{71} \equiv 2^{71 \bmod 4} \equiv \boxed{8}$

Let

$S(n)=(1+\sqrt{71})^n+(1-\sqrt{71})^n\dots(1)$

We want to find the last digit of $S(71)$ .

My strategy is to find the quadratic equation whose roots are the two bracketed terms in (1), and then to examine the pattern of the sums of the powers of the roots of this equation to find $S(71)$ .

The required quadratic equation is

$\left(x-(1+\sqrt{71})\right)\left(x-(1-\sqrt{71})\right)=0$

$\implies x^2-2x-70=0$

Now invoking Newtons marvellous iterative identities for the sums of the roots of a polynomial -

$S(1)=2$

$S(2)=2\times2+70\times2=144$

$S(3)=2\times 144 +70\times2$

and so on.

Each further iteration involves multiplication by two (and the addition of a multiple of 70 which does not change the last digit!).

The final digits thus cycle through the pattern

$2,4,8,6,2,4,8,6,\dots$

and so the final digit of $S(71)$ is

$2^{71\bmod 4}=2^3=8$

S (10) = 10(1 - 1/2)(1 - 1/5) = 4 .......... (FERMAT'S THEOREM)

Let square root of 71 be denoted by x

Then the given expression =

(x + 1)^71 - (x - 1)^71 = (x + 1)^3 - (x - 1)^3 (mod 10)

x = square root of 71 = square root of 1 (mod 10)

Then

(x + 1)^71 - (x - 1)^71 = 2^3 - 0 = 8 (mod 10)

So , the last digit of the given number is 8

According to the theory of Binet Forms, the sequence $x_n=(1+\sqrt{71})^n+(1-\sqrt{71})^n$ satisfies the recursive equation $x_n=2x_{n-1}+70x_{n-2}$ , with $x_0=x_1=2$ . Thus $x_{n}\equiv2x_{n-1} \pmod{10}$ and $x_{n}\equiv2^n \pmod{10}$ . Now $x_{71}\equiv2^{71}\equiv2^3=\boxed{8} \pmod{10}$

We have

$(\sqrt{71})+1)^{71}-(\sqrt{71})-1)^{71}=(\sqrt{71}+1)^{71}+(1-\sqrt{71})^{71} = 2\sum_{m=0}^{35}{{71}\choose{2m}}71^{m}$

because the odd terms cancel out and the even one sum up. So we have to find these expression $\pmod{10}$ . Write:

$2\sum_{m=0}^{35}{{71}\choose{2m}}71^{m}=2\sum_{m=0}^{35} {{71}\choose{2m}}(70+1)^{m} =2\sum_{m=0}^{35} {{71}\choose{2m}}\sum_{k=0}^{m}{{m}\choose{k}}7^{k}10^{k}$

and of course the only term in the second sum that survives after taking the congruence $\pmod{10}$ is the $k=0$ term, so:

$2\sum_{m=0}^{35} {{71}\choose{2m}}71^{m} \equiv 2\sum_{m=0}^{35} {{71}\choose{2m}} \pmod{10}$ .

Consider now the identity

$2^{71}=(1+1)^{71}+(1+(-1))^{71}=\sum_{k=0}^{71} {{71}\choose{k}}+\sum_{k=0}^{71} {{71}\choose{k}}(-1)^{k}=2\sum_{m=0}^{35} {{71}\choose{2m}}$

and so we finally have that

$(\sqrt{71}+1)^{71}-(\sqrt{71}-1)^{71} \equiv 2^{71} \equiv 2*[2^{10}]^{7} \equiv 2*[4]^{7} \equiv 8 \pmod{10}$

Define $F_n = (1 + \sqrt{71} )^n + (1 - \sqrt{71} )^n.$ Observe that we want the last digit of $F_{71}$ . Observe that the characteristic polynomial of $F_n$ is $x^2 - 2x - 70$ , so we must have that $F_n = 2F_{n-1} + 70F_{n-2}.$ From here we can easily see that $F_n \equiv 2F_{n-1} \pmod{10}$ . Hence, the unit digits cycle $2,4,8,6$ . Since $71 \equiv 3 \pmod 4$ , the answer is $\boxed{8}$ .

First of all I thank to problem writer :)

(71.1/2 + 1)^71 - (71.1/2 -1)^71

= (71/2+1)^71 - (71/2-1)^71

=((71+2)/2))^71 - ((71-2)/2)^71

=(73/2 - 69/2)^71

=((73-69)/2)^71

=(4/2)^71

=2^71

Now we know how to find last digit of any number. If we mod by 10 on any number we got last digit of that number.So now 2 multiply by itself up to 71 times and every time we mod by 10 and finally we got last digit of the equation and that is 8.