MHB No Integer x for Which $P(x)=14$ Given Four Integer Values of $P(x)=7$

[kaliprasad]

Show that if a polinomial $P(x)$ with integer coefficients takes the value 7 for four different integer values of x then there is no integer x for which $P(x) = 14$

 

[Albert1]

Show that if a polinomial $P(x)$ with integer coefficients takes the value 7 for four different integer values of x then there is no integer x for which $P(x) = 14$

Spoiler

let $P(x)=Q(x)(x-a)(x-b)(x-c)(x-d)+7$
that is $P(a)=P(b)=P(c)=P(d)=7$
where $a,b,c,d$ are four different integer values
and $P(x),Q(x) $ with integer coefficients
if e is another integer value and $P(e)=14$ then we have
$P(e)-7=7=Q(e)(e-a)(e-b)(e-c)(e-d)----(1)$
where $Q(e),(e-a),(e-b),(e-c),(e-d)\in Z$
this is impossible for 7 is a prime ,and the proof is done
(1) can be true if $Q(e)\in Q$

 

[kaliprasad]

Spoiler

P(x) -7 is zero at 4 different values say a,b,c,d

hence
$P(x)-7 = Q(x)(x-a)(x-b)(x-c)(x-d)$

or
$P(x)= Q(x)(x-a)(x-b)(x-c)(x-d)+7 $

Let

$P(t) = 14 = Q(t)(t-a)(t-b)(t-c)(t-d)+7 $

hence $Q(t)(t-a)(t-b)(t-c)(t-d)= 7 $

so 7 must have 5 factors out which 4 are different. but 7 cannot have 2 different factors (7 * -1 * -1 ) or (7 * 1) so above relation cannot be true so there is no solution