if-q-is-prime-number-fixed-then-solve-for-natural-numbers-the-equation-1-q-1-x-1-y-1-z-

Question Number 149567 by mathdanisur last updated on 06/Aug/21

i f q i s p r i m e n u m b e r f i x e d, t h e n

s o l v e f o r n a t u r a l n u m b e r s t h e e q u a t i o n :

\frac{1}{q} = \frac{1}{x} + \frac{1}{y} - \frac{1}{z}

Commented by Rasheed.Sindhi last updated on 07/Aug/21

My answer is too lengthy .

The question is waiting for a better

solution .

Commented by mathdanisur last updated on 07/Aug/21

Dear S er

We hawe : \frac{1}{n} = \frac{1}{2 (n - 1)} + \frac{1}{2 (n + 1)} - \frac{1}{(n - 1) (n + 1)}

\Rightarrow n = q \dots .

Ser, this true or wrong . ?

Commented by Rasheed.Sindhi last updated on 07/Aug/21

\frac{1}{2 (q - 1)} + \frac{1}{2 (q + 1)} - \frac{1}{(q - 1) (q + 1)}

= \frac{q + 1 + q - 1 - 2}{2 (q - 1) (q + 1)} = \frac{2 q - 2}{2 (q - 1) (q + 1)}

= \frac{2 (q - 1)}{2 (q - 1) (q + 1)} = \frac{1}{q + 1} \neq \frac{1}{q}

N o t c o r r e c t s e r .

Commented by mathdanisur last updated on 07/Aug/21

Sorry S er, \dots \frac{1}{n (n - 1) (n + 1)}

Commented by Rasheed.Sindhi last updated on 07/Aug/21

\frac{1}{2 (n - 1)} + \frac{1}{2 (n + 1)} - \frac{1}{n (n - 1) (n + 1)}

\frac{n (n + 1) + n (n - 1) - 2}{2 n (n - 1) (n + 1)}

\frac{n^{2} + n + n^{2} - n - 2}{2 n (n - 1) (n + 1)} = \frac{{2 n}^{2} - 2}{2 n (n - 1) (n + 1)}

\frac{2 (n - 1) (n + 1)}{2 n (n - 1) (n + 1)} = \frac{1}{n} ✓

It means that this is the solution .