Mr. prime!

Number Theory Level 3

$\begin{aligned} & 5^2 + 7^2 = 74\equiv \pm 2\pmod 4 \\& 13^2 + 23^2 = 698 \equiv \pm 2 \pmod 4 \\& 103^2 + 101^2 = 20810\equiv \pm 2\pmod 4\end{aligned}$

Mr.prime takes any two distinct prime numbers $p_1$ and $p_2$ greater than 2 . He sums the square of primes (as above) and found that an even number leaves the remainder of $\pm 2$ when divided by $4$ .

Is it true that the numbers created in such way always leaves remainder $\pm2$ when divided by $4$ ?

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No Yes

1 solution

Naren Bhandari
May 9, 2018

Primes are odd numbers which are called odd primes greater than 2. Therefore, let $p_1 = 2n_1\pm 1$ and $p_2 =2n_2\pm 1$ where $n_1$ and $n_2 > 0$ . Now $p_1^2 + p_2^2 = (2n_1\pm 1 )^2 + (2n_2\pm 1)^2 \\ e = 4n_1^2 \pm 4n_1 + 4n_2^2 \pm 4n_2 + 2= 4\,(n_1^2+n_2^2 \pm n_1\pm n_2 )+2 \\ \text{OR} \\ e= 4n_1^2 \pm 4n_1 + 4n_2^2 \pm 4n_2 +{\color{#3D99F6}4-2} = 4\, (n_1^2+n_2^2\pm n_1\pm n_2+1)-2$ shows that when even number $e$ is divisible by $4$ leaving the remainder $\pm 2$ . Hence it is true .

The square of an odd number divided by 4 lefts a remainder 1,so 1+1=2

X X - 3 years, 1 month ago

Yeah!! That true, I also observe that while solving however, I was interested in that way. :)

Naren Bhandari - 3 years, 1 month ago

Mr. prime!

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