Huge powers

Number Theory Level 3

First two consecutive prime numbers greater than $10^7$ are $p_{1} = 10000019$ and $p_{2} = 10000079$ respectively. Let $p$ be the smallest prime factor of $S = p_{1}^2 + p_{2}^2-5$ . Also, let $q$ be the units digit of $S$ . What is the value of $p+q$ ?

The answer is 10.

1 solution

Snehal Shekatkar
Jul 18, 2016

Since $p_1$ is prime, it is not divisible by $3$ . Hence, either $p_1 - 1$ or $p_1+1$ is divisible by 3. This means $(p_1-1)(p_1+1)=p_1^2-1$ is certainly divisible by $3$ (One could use the Fermat's theorem to reach the same conclusion). Similar argument holds for $p_2$ . Hence we have,

$p_{1}^2+p_2^2 \equiv 2 (\text{mod} 3)$

$\therefore p_{1}^2+p_2^2 \equiv 5 (\text{mod} 3)$

Hence $S$ has $p = 3$ as its smallest prime factor.

Finding units digit is easy. Since both numbers end with $9$ , their squares would end with $1$ and sum of their squares would end with $2$ . Hence the units digit of $S$ would be $q=7$ .

Hence, $p+q=\boxed{10}$

Moderator note:

Note that we need to show that the expression is not a multiple of 2, in order to conclude that the smallest prime factor is 3.

Huge powers

The answer is 10.

1 solution

Moderator note:

0 pending reports