[Guest]

G(x) is a nonconstant polynomial in x with integer coefficients and, there exist five distinct integers a₁, a₂, a₃, a₄, a₅ such that:

G(a₁)= G(a₂)= G(a₃)= G(a₄)= G(a₅)= 2.

Prove that there does not exist any integer b, such that: G(b)= 9.

What is the minimum value of a nonzero perfect square that G(b) can equal? How about the minimum number of the form 3^C that G(b) equals, where C is a positive integer?

Edited October 6, 2009 by K Sengupta

[Guest]

Given the problem, G(x) seems to be of the form

G(x) = f(x) + c

where f(x) contains terms of coefficients of x only

On closer look, it is of the form

G(x) = (x-a1)(x-a2)(x-a3)(x-a4)(x-a5) + 2

Then, for G(x) to be 9,

(x-a1)(x-a2)(x-a3)(x-a4)(x-a5) = 7

Since the integer factors (including negatives) are only 1,-1,+/-7

a product of 5 integers can not be 7

So, there can be no value of x for which G(x) may be 9

For minimum value of a perfect square, (x-a1)(x-a2)(x-a3)(x-a4)(x-a5) = N - 2

where N is a perfect square and N-2 has at least 5 factors (including negatives)

The minimum possible perfect square is 100 where 100-2 = 98 can have factors of 1,-1,2,-7,7

of the three factors one is a negative i.e. either -2,7,7 or 2,-7,7

For minimum number of the form 3^c

(x-a1)(x-a2)(x-a3)(x-a4)(x-a5) = R - 2

where R is of the form 3^c and R-2 has at least 5 factors (including negatives)

This is possible for a minimum vlaue of 3⁷ where 3⁷ = 2187 and 2187-2 = 2185 has 5 distinct factors (1,-1,-5,23,19)

of the three 5,23 & 19 any one of these can have a negative vlaue

Edited October 7, 2009 by DeeGee

[Guest]

You don't say the order of the polynomial, the assumption was 5th order but might a higher order polynomial work?

[Guest]

My impression of KS's question is that whatever the order is of the polynomial concerned, his final statement must be false. A fifth order polynomial with 5 initial conditions is too constricting. Oh yes, where does he get all these problems from?

Edited October 8, 2009 by jerbil

[Guest]

You don't say the order of the polynomial, the assumption was 5th order but might a higher order polynomial work?

Well you could always write G(x) as

G(x) = (x-a1)^l(x-a2)^m(x-a3)ⁿ(x-a4)^o(x-a5)^p + 2

Get whatever degree of the polynomial you like!

[Guest]

Well you could always write G(x) as

G(x) = (x-a1)^l(x-a2)^m(x-a3)ⁿ(x-a4)^o(x-a5)^p + 2

Get whatever degree of the polynomial you like!

But the OP doesn't state that a1 to a5 are the *only* integers for which G(x) = 2. So G(x) can also be:

G(x) = (x-a1)^l(x-a2)^m(x-a3)ⁿ(x-a4)^o(x-a5)^p(x-b1)^r1..(x-bn)^rn + 2

[Guest]

But the OP doesn't state that a1 to a5 are the *only* integers for which G(x) = 2. So G(x) can also be:

G(x) = (x-a1)^l(x-a2)^m(x-a3)ⁿ(x-a4)^o(x-a5)^p(x-b1)^r1..(x-bn)^rn + 2

Doesn't DeeGee's analysis still hold true for this G(x)?

[Guest]

Doesn't DeeGee's analysis still hold true for this G(x)?

It does, sorry for throwing that in.