prove-or-disprove-i-1-n-p-1-i-p-2-p-1-p-2-P-1-2-np-1-n-1-p-2-1-2-n-n-1-p-2-p-1-n-2-n-2-p-1-p-2-p-i-a-1-a-2-a-n-a-i-1-2-p-1-Z-p-1-0-n-Z-n-2-n-k-0-k-2-p-

Question Number 3794 by Filup last updated on 21/Dec/15

prove or disprove : \underset{i = 1}{\sum^{n}} p_{1} i = p_{2}

p_{1}, p_{2} \in P

\frac{1}{2} {n p}_{1} (n + 1) = p_{2}

\frac{1}{2} n (n + 1) = \frac{p_{2}}{p_{1}}

n^{2} + n = \frac{2}{p_{1}} p_{2}

p_{i} \neq a_{1} a_{2} \dots a_{n} a_{i} \neq 1

∴ \frac{2}{p_{1}} \notin Z p_{1} \neq 0

n \in Z

n^{2} + n - k = 0

k = \frac{2}{p_{1}} p_{2}

∴ n = \frac{- 1 \pm \sqrt{1 + 4 k}}{2}

∴ n = \frac{- 1 \pm \sqrt{1 + \frac{8}{p_{1}} p_{2}}}{2}

∴ n \notin Z

∴ \underset{i = 1}{\sum^{n}} p_{1} i \neq p_{2}

Is this correct ? Or am I wrong ?

Commented by prakash jain last updated on 21/Dec/15

\underset{i = 1}{\sum^{n}} p_{1} i = X = \frac{n (n + 1) p_{1}}{2}

p_{1} divides X and \frac{n (n + 1)}{2} is

an integer (> 1 for n > 1) divides X .

So X \notin P for n > 1 .

Commented by prakash jain last updated on 21/Dec/15

Your rational is also correct since if

\frac{n (n + 1)}{2} p_{1} is a prime than n is not an integer .

so \frac{n (n + 1)}{2} p_{1} \notin P for \forall n \in N and n > 1

Commented by Yozzii last updated on 21/Dec/15

s = \underset{i = 1}{\sum^{n}} i g i v e s t r i a n g u l a r n u m b e r s a n d

h e n c e s \in N . N o w, p_{2} \in P a n d p_{1} \in P .

T h e r e f o r e, {s p}_{1} = p_{2} i s t r u e i f f s = 1

a n d p_{1} = p_{2} . O t h e r w i s e, {s p}_{1} \neq p_{2} s i n c e

p_{2} = p_{1} s i s s u g g e s t i v e o f p_{2} \notin P i f s > 1

o r t h e g i v e n e q u a t i o n i s a n i n e q u a t i o n

i f p_{1} \neq p_{2} a n d s ⩾ 1 .

Commented by Yozzii last updated on 21/Dec/15

I f p_{1} = p_{2},

n = \frac{- 1 \pm \sqrt{9}}{2} = \frac{\pm 3 - 1}{2} = 1, - 2

n \in N \Rightarrow n = 1 o n l y .

T h i s s h o u l d b e a d d e d t o t h e p r o o f

I t h i n k .

Commented by Filup last updated on 21/Dec/15

Thank you!

Facebook Tweet Pin