prove-any-prime-number-gt-2-can-be-written-into-x-2-y-2-where-x-y-N-

Question Number 8027 by Nayon last updated on 28/Sep/16

p r o v e \to a n y p r i m e n u m b e r > 2

c a n b e w r i t t e n i n t o (x^{2} - y^{2}) w h e r e

(x, y) \in N

Commented by FilupSmith last updated on 28/Sep/16

Attempt to solve \dots currently playing

around with math \dots

(x, y) \in N

P \in P / {2} {P = 3, 5, 7, 11, 13, \dots}

x^{2} - y^{2} = P

P > 0 \Rightarrow x > y

x = \underset{i = 1}{\prod^{ω (x)}} p_{i}^{a_{i}} y = \underset{i = 1}{\prod^{ω (y)}} q_{i}^{b_{i}}

x = p_{1}^{a_{1}} p_{2}^{a_{2}} p_{3}^{a_{3}} \dots p_{n}^{a_{n}} (e . g . 150 = 15 \times 10 = 5 \times 3 \times 5 \times 2

= 2^{1} \cdot 3^{1} \cdot 5^{2}

ω (x) = n (number of discrete prime factors)

∴ if x^{2} - y^{2} = P \overset{?}{\Rightarrow} \exists p_{k} \in (x \lor y) : p_{k} = P

Answered by prakash jain last updated on 28/Sep/16

x^{2} - y^{2} = (x - y) (x + y)

If p = x^{2} - y^{2}

p can have only 2 factors (1, p)

x - y = 1

x + y = p

x = \frac{p + 1}{2}

y = \frac{p - 1}{2}

All prime > 2 are odd can be

written as 2 k + 1

p = 2 k + 1 \Rightarrow x = k + 1, y = k

\Rightarrow x, y \in N

p = 2 k + 1 = {(k + 1)}^{2} - k^{2}

Commented by FilupSmith last updated on 29/Sep/16

i forgot x^{2} - y^{2} = (x - y) (x + y) h a h a