Question

let f(x) be a quadratic polynomial. if, f(0),f(1),f(-1)\(\in\mathbb{z}\), then show that f(n)\(\in\mathbb{z}\) for all \(n\in\mathbb{z}\)

Answer 1

By PMI ?

Answer 2

what's that?

Answer 3

Principle of Mathametical Induction

Answer 4

let f(x) = ax^2 + bx + c

Answer 5

Sadly, mathematical induction holds only for natural numbers, not all integers..

Answer 6

let n1,n2,n3 be some integers

f(x) = n1 + n2x + n3x(x+1)

f(0) = n1
f(-1) = n1 - n2
f(1) = n1 + n2 + 2n3

satisfies the conditions

Answer 7

i spose n was a bad choice for genericy ...

Answer 8

show that for all z, z n2 and z(z+1)n3 are integers and closed

Answer 9

or in other words: Z is closed under addition and multiplication therefore z n2 is an integer; and z(z+1) n3 is an integer

Answer 10

let a,b,c,n in Z; and let f(x) = a + bx + cx(x+1) be a quadratic polynomial

f(0) = a is an integer
f(-1) = a - b is an integer since Z is closed under addition/subtraction
f(1) = a + b + 2c is an integer since Z is closed under addition and multipication

let x = n

f(n) = a + bn + c n(n+1) is an integer since Z is closed .....

Answer 11

Does, this not only prove for integral quadratic polynomial? can't there be other cases too where a,b,c are not integers, yet f(0), f(1), f(-1) are integers?

Answer 12

thats not what it is asking for .....

Answer 13

let f(x) be a quadratic poly ... NOT, let f(x) be all quadratic polys

Answer 14

I understand your solution but here you are taking an integral quadratic question which is just a special case of quadratic equation.

Answer 15

spose we introduce a real number, d such that

f(x) = a + bx + cx(x+1) + dx(x+1)(x-1)

then for f(-1,0,1) we have integer solutons; but we cannot guarentee integer solutions for any other "n" in Z

Answer 16

i am taking a quadratic polynomial that has integer solutions for x= -1,0,1

Answer 17

lol, i turned d into a cubic

Answer 18

this can still be proved by using equation
ax^2 +bx +c
can't it be?

Answer 19

i see what your trying to say tho; you think that my f(x) is not general enough

Answer 20

i do not believe that it can be proved in that manner ... at least nothing comes to mind

c would have to be an integer

and a and b would still have to be integers when x=1 and x=-1; unless you can find some non integer values that add up to an integer (some fractions perhaps, but not all fractions work)

Answer 21

ok, supposing a,b,c are integers and f(x)=ax^2+bx+c
we have
f(0)=c
f(1)=a+b+c
f(-1)=a-b+c
which are all integers
and again,
n is an integer, so,
f(n)=an^2+bn+c is an integer

Is there any special reason why you choose that particular equation?

Answer 22

if a and b are fractions; then they would have to satisfy some b = z-a relationship maybe

p/q n^2 + s/t n

n (p/q n + s/t)

p/q n + s/t = z

p/q = (z - s/t)/n

Answer 23

i simply built it so that each term is zeroed out is all

Answer 24

So, it holds true for any arbitrary quadratic equation as long as all co-efficients are integers?

Answer 25

even in the case where you could use rational numbers for a and b there is still a limited set of fraction that would be available to use and therefore not be "generic" enough to prove ALL poly quads that have f(-1,0,1) in Z

Answer 26

yes, f(n) = an^2 + bn + c; would by default have all integer solutions if abc in Z

Answer 27

\[[f(n)=\frac pqn^2+\frac stn+c]\in Z,~if~(\frac pq=\frac{k-(s/t)}{n})~for~k\in Z, n\ne0\]

Answer 28

I guess, i will go with the above solution! :D
and stop thinking about fractions..

Answer 29

to find all possible cases: let a,b,c in Z such that

0 0 1 a
1 1 1 b
1 -1 1 c

row reduces to give us something like:\[f(x)=\frac{b+c-2a}{2}x^2+\frac{b-c}{2}x+a\]

Answer 30

This one seems more complicated..

Answer 31

it is, since we are considering all cases of f(x) that would give us integers for f(-1,0,1); if for all bc in Z this pans out then we have a proof

Answer 32

if: \[\frac{b+c-2a}{2}n+\frac{b-c}{2}=k~:~k\in Z\]
we will have our proof

Answer 33

why do we need to find such cases? It is clearly mentioned in question that f(0), f(1), f(-1) are integers..

Answer 34

and when they are integers, we have to prove f(x) is an integer too for all x belongs to R

Answer 35

yes, it says that the results are integers; so letting abc be integer solutions we are able to form a matrix that we can row reduce to determine all suitable values for the coefficients that will produce the results

Answer 36

Z*

Answer 37

since a quad poly is uniquely defined by 3 points, we are able to construct all possible f(-1,0,1) results in this manner

Answer 38

I don't think this question is that complicated.. May be the above method is sufficient!
Thanks!

Answer 39

\[\frac{b+c-2a}{2}n+\frac{b-c}{2}=k\]
\[n(b+c-2a)+b-c=2k\]
\[b(n+1)+c(n-1)-2n~a=2k\]
since all we have remaining are integer products and sums, that should be proofed