Question 926307


if {{{x^2-63x-64=0}}} and {{{p}}} and {{{n}}} are integers such that {{{p^n=x}}}, then


{{{(p^n)^2-63p^n-64=0}}}

{{{p^2n-63p^n-64=0}}}.......use quadratic formula

{{{p^n = (-(-63) +- sqrt( (-63)^2-4*1*(-64) ))/(2*1) }}}

{{{p^n = (63 +- sqrt(3969+256 ))/2 }}}

{{{p^n = (63 +- sqrt(4225 ))/2 }}}

{{{p^n = (63 +- 65)/2 }}}

solutions:

{{{p^n = (63 + 65)/2 }}}

{{{p^n = 128/2 }}}

{{{p^n = 64 }}}

or

{{{p^n = (63-65)/2 }}}

{{{p^n =-1/2 }}}




solutions for  variable {{{p}}}

{{{p = (-1)^(1/n)}}}

{{{p = 64^(1/n)}}}

so, when {{{p = -1}}} we have an integer solution

{{{(64^-1)^2-63*64^-1-64=0}}}

{{{(1/64)^2-63*(1/64)-64=0}}}

{{{1/4096-63/64-64=0}}}

{{{0.000244140625-0.984375-64=0}}}

{{{-0.984130859375-64=0}}}


{{{-64.984130859375<>0}}}; so, {{{highlight(p = -1)}}} cannot be a value for {{{p}}} 
because gives us {{{-64.984130859375<>0}}}


{{{p}}} can be {{{64}}} because {{{  n = 1}}} and

{{{(64^1)^2-63*64^1-64=0}}}

{{{(64^1)^2-63*64^1-64=0}}}

{{{4096-4032-64=0}}}

{{{4096-4096=0}}}

{{{0=0}}}