IIT JEE , AIEEE Forum - Mathematics , Algebra

aforadi

let p(x) be a cubic polynomial with integral coefficients . also a,b,c are integers such that p(a)=b,p(b)=c,p(c)=a. find no of cubic polynomials (integral coeff.) which satisfy this condition.
let p(x) be a polynomial with integral coefficients such that p(0) and p(1)
are odd integers. find no of soln of p(x)=0

faith

1st ques . infinte , as four variables n three eqn can b solved in many ways
2nd ques , min. 0 real roots

aforadi

both are wrong

aforadi

in the 2nd qn i want the exact ans not min.

faith

then u need to give the exact degree of d polynomial

aforadi

the result wld be true for all polynomial of any degree

anit_sahu

assuming a,b,c to be distinct no integral coefficient ploynomial can exist.
LEMMA-(x-y)divides p(x)-p(y)
proceeding we get +-(a-b)=+-(b-c)=+-(c-a) (+- means plus minus)
manipulating this we get a,b,c are equal which is impossible.so no polynomial exists
for 2nd question
LEMMAif a is an integer root of f(x),(a-m) divides f(m)
if a is an integer root of p(x) then a not equal to 0,1.a must divide f(0) according to lemma and a must be odd to satisfy it.so taking m=1 in the lemma we see that an even number (a-1) divides the odd number f(1) which is an contradiction.so the polynomial has got no integer roots

aforadi

cld u pls explain lemma

anit_sahu

which lemma do u want to be explained

anit_sahu

ok im explaining both of them.
ist lemma
let f(x)=ax^2+bx+c
then f(y)=ay^2+by+c
f(x)-f(y)=a(x+y)(x-y)+b(x-y)=(x-y)(a(x+y)+b)
hope this is clear.can be easily concieved for higher powers
2nd lemma
by remainder theorem
f(x)=(x-m)q(x)+f(m)
let a be the integer root
0=f(a)=(a-m)q(a)+f(m) or f(m)=-(a-m)q(a)
hence (a-m) divides f(m)

iitkgp_bipin

Anit sahu has explained it correctly.

Still if there is any doubt do ask.

faith

"proceeding we get +-(a-b)=+-(b-c)=+-(c-a) (+- means plus minus)"

how do we get this ?

nwhats "LEMMA" ?

aforadi

thnx a lot

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