Problem no. 25 ( IMO 1993) Let f(x) = , where is n integer . Prove that f(x) cannot be expressed as the product of non constant polynomial with integer coefficients .

SOLUTION:(by Neelanjan Mondal )

Suppose

f(x) = . We show that all the a’s are divisible by 3 and using that we will establish a contradiction .[Here , we will show using only + ve sign of 3 , the other case is

f(x) = , in which constant terms contains only – ve terms and the casee follows only the + ve case . So , we will show here only the first case] .

r and s must be greater than 1.Because

for if r = 1, then 3 is a root . Now we will make two cases :

case 1:

n is even, then it follows that 0 = , which is false since 3+5 = 8.

case 2 :

if n is odd we would have 0 = , which is false since 3 + 5 = 8.

If s = 1, then 1 is a root and we will argue by contradiction in the same way .

So rn – 2, and hence the coefficients of x, are all zero. Since the coefficient of x is zero, we have: , so is divisible by 3.

Now , we will use the induction hypothesis .

We can

now proceed by induction.

Assume are all divisible by 3. Then

consider the coefficient of . If s-1 t+1, then = linear combination of . If s-1 t+1, then = linear combination of some or all of

Either way, is divisible by 3.

Now we will consider the coefficients of x, ,

… , which gives us that all the a’s are multiples of 3. Now consider the coefficientof which is also zero.

Now we will get , that is a sum of of terms which are multiples of 3 .

Then it does not becomes 0.

So , follows a contradiction !

So , the factorization is not possible .( proved ).

# IMO 1993

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