Suppose-an-integer-x-a-natural-number-n-and-a-prime-number-p-satisfy-the-equation-7x-2-44x-12-p-n-Find-the-largest-value-of-p-

Question Number 21200 by Tinkutara last updated on 15/Sep/17

Suppose an integer x, a natural

number n and a prime number p

satisfy the equation 7 x^{2} - 44 x + 12 = p^{n} .

Find the largest value of p .

Commented by mrW1 last updated on 16/Sep/17

47 ?

Answered by alex041103 last updated on 16/Sep/17

7 x^{2} - 44 x + 12 = (x - 6) (7 x - 2) = p^{n}

\Rightarrow x - 6 = p^{n_{1}} a n d 7 x - 2 = p^{n_{2}} a n d n_{1} + n_{2} = n

\Rightarrow x = p^{n_{1}} + 6 a n d 7 p^{n_{1}} + 40 = p^{n_{2}}

\Rightarrow n_{1} < n_{2}

\Rightarrow 7 + \frac{40}{p^{n_{1}}} i s a w h o l e n u m b e r

\Rightarrow p^{n_{1}} i s a d i v i s o r o f 40 = 2^{3} \times 5^{1} \times {(s o m n u m)}^{0}

(p, n_{1}) = (p, 0), (5, 1), (2, 3)

N o w w e t r y p = 5 o r 2 a n d i t {d o e s n}^{'} t w o r k

\Rightarrow p^{n_{2}} = 7 \times p^{0} + 40 = 47 (47 i s p r i m e)

\Rightarrow p = 47, n_{1} = 0, n_{2} = 1, n = 1

A n s . p_{m a x} = p = 47 a n d

7 x^{2} - 44 x + 12 =

= 7 \times 7^{2} - 44 \times 7 + 12 = 47^{1} = p^{n}

Commented by Tinkutara last updated on 16/Sep/17

Thank you very much Sir!