## IMC 2011 Day 2 Problem 4

Let be a polynomial with real coefficients of degree . Suppose that is an integer for all . Prove that for all distinct integers .

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Let be a polynomial with real coefficients of degree . Suppose that is an integer for all . Prove that for all distinct integers .

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Let be a polynomial in of degree , where each belongs to . It is given that is an integer for , So (say) _ _ _ _ (1), this yields that . (By substituting in equation (1) and equating the corresponding co-efficient) _ _ _ _ (2) and note that y lies between and . Equation (1) conveys, whenever lies between and , the curve reduces to a straight line.

Now consider the expression , let’s assume that ‘’ be any integer but be an integer such that lies between and . ( even though it is a special case, first we will prove the problem for this case, which in turn leads to the solution when both a and b are any integers.)

We will take induction on , whenever the result is trivially true. Suppose be an integer ‘’ for all integers less than or equal to , using equation (2) and expanding , equating the corresponding co-efficient we will get _ _ _ (3)( note that it is true for all integer less than or equal to ). Equation (3) says that a polynomial of degree can be written as a polynomial of degree for all integers less than or equal to a for any positive integer . Now consider , on expanding and using equation (3) we will get , therefore is an integer. This closes the induction and proves that divides for any integer a and an integer lies between and .

Finally consider any distinct integers and , can be written as , where lies between and , therefore and , from the last paragraph. Therefore divides .

Hence the proof.

I don’t think your proof is correct. The problem says that the ratio is AN integer for every between and , not THE same integer. This means that for every there can be a different .

I don’t think your proof is correct. The problem says that the ratio is AN integer for every between and , not THE same integer. This means that for every there can be a different .

Yea, i know that but whether its same integer or different, the idea is that curve is reduced to a straight line in that range and that is the main point in the proof.

( Readers, here i am not able to type powers and subscripts so i assume that you will make out)

Let f(p) = anpn + an-1pn-1+……..+a0 be a polynomial in p of degree n, where each ai belongs to R. It is given that f(x) – f(y) / x-y is an integer for 0≤x<y≤n, So { f(x) – f(y) / x-y} = k ( say) _ _ _ _ (1), this yields that f(y) = ky + a0. (By substituting x = 0 in equation (1) and equating the corresponding co-efficient) _ _ _ _ (2) and note that y lies between 0 and n. Equation (1) conveys, whenever x lies between 0 and n, the curve f(x) reduces to a straight line.

Now consider the expression {f(a) – f(b)} / a-b, let’s assume that ‘a’ be any integer but b be an integer such that b lies between 0 and n. ( even though it is a special case, first we will prove the problem for this case, which in turn leads to the solution when both a and b are any integers.)

We will take induction on a, whenever a = 0 the result is trivially true. Suppose {f(a) – f(b)} / a-b be an integer ‘m’ for all integers less than or equal to a, using equation (2) and expanding f(a), equating the corresponding co-efficient we will get f(a) = ka + a0 _ _ _ (3)( note that it is true for all integer less than or equal to a). Equation (3) says that a polynomial of degree n can be written as a polynomial of degree 1 for all integers less than or equal to a for any positive integer n. Now consider f(a+1) = an(a+1)n + …..+ a0, on expanding and using equation (3) we will get f(a+1) = k(a+1) + a0, therefore {f(a+1) – f(b)}/ (a+1) – b is an integer. This closes the induction and proves that (a – b) divides f(a) – f(b) for any integer a and an integer b lies between 0 and n.

Finally consider any distinct integers a and c, f(a) – f(c) can be written as {f(a) – f(b)} – { f(c) – f(b)}, where b lies between 0 and n, therefore f(a) – f(b) = K(a – b) and f(c) – f(b) = K(c – b), { from the last paragraph}. Therefore a – c divides f(a) – f(c).

Hence the proof.

Even if you write different integer the proof remains the same.

Yea, i know that but whether its same integer or different, the idea is that curve is reduced to a straight line in that range and that is the main point in the proof.

( Readers, here i am not able to type powers and subscripts so i assume that you will make out)

Let f(p) = anpn + an-1pn-1+……..+a0 be a polynomial in p of degree n, where each ai belongs to R. It is given that f(x) – f(y) / x-y is an integer for 0≤x<y≤n, So { f(x) – f(y) / x-y} = k ( say) _ _ _ _ (1), this yields that f(y) = ky + a0. (By substituting x = 0 in equation (1) and equating the corresponding co-efficient) _ _ _ _ (2) and note that y lies between 0 and n. Equation (1) conveys, whenever x lies between 0 and n, the curve f(x) reduces to a straight line.

Now consider the expression {f(a) – f(b)} / a-b, let’s assume that ‘a’ be any integer but b be an integer such that b lies between 0 and n. ( even though it is a special case, first we will prove the problem for this case, which in turn leads to the solution when both a and b are any integers.)

We will take induction on a, whenever a = 0 the result is trivially true. Suppose {f(a) – f(b)} / a-b be an integer ‘m’ for all integers less than or equal to a, using equation (2) and expanding f(a), equating the corresponding co-efficient we will get f(a) = ka + a0 _ _ _ (3)( note that it is true for all integer less than or equal to a). Equation (3) says that a polynomial of degree n can be written as a polynomial of degree 1 for all integers less than or equal to a for any positive integer n. Now consider f(a+1) = an(a+1)n + …..+ a0, on expanding and using equation (3) we will get f(a+1) = k(a+1) + a0, therefore {f(a+1) – f(b)}/ (a+1) – b is an integer. This closes the induction and proves that (a – b) divides f(a) – f(b) for any integer a and an integer b lies between 0 and n.

Finally consider any distinct integers a and c, f(a) – f(c) can be written as {f(a) – f(b)} – { f(c) – f(b)}, where b lies between 0 and n, therefore f(a) – f(b) = K(a – b) and f(c) – f(b) = K(c – b), { from the last paragraph}. Therefore a – c divides f(a) – f(c).

Hence the proof.

Even if you write different integer the proof remains the same.

Yea, i know that but whether its same integer or different, the idea is that curve is reduced to a straight line in that range and that is the main point in the proof.

( Readers, here i am not able to type powers and subscripts so i assume that you will make out)

Let f(p) = anpn + an-1pn-1+……..+a0 be a polynomial in p of degree n, where each ai belongs to R. It is given that f(x) – f(y) / x-y is an integer for 0≤x<y≤n, So { f(x) – f(y) / x-y} = k ( say) _ _ _ _ (1), this yields that f(y) = my + a0. (By substituting x = 0 in equation (1) and equating the corresponding co-efficient) _ _ _ _ (2) and note that y lies between 0 and n. Equation (1) conveys, whenever x lies between 0 and n, the curve f(x) reduces to a straight line.

Now consider the expression {f(a) – f(b)} / a-b, let’s assume that ‘a’ be any integer but b be an integer such that b lies between 0 and n. ( even though it is a special case, first we will prove the problem for this case, which in turn leads to the solution when both a and b are any integers.), let f(b) = qb + a0.

We will take induction on a, whenever a = 0 the result is trivially true. Suppose {f(a) – f(b)} / a-b be an integer ‘r’ for all integers less than or equal to a, using equation (2) and expanding f(a), equating the corresponding co-efficient we will get f(a) = qa + a0 _ _ _ (3),HERE THE HIGHLIGHT IS CO-EFFICIENT OF a IN f(a) IS SAME AS THAT OF b IN f(b). (note that it is true for all integer less than or equal to a). Equation (3) says that a polynomial of degree n can be written as a polynomial of degree 1 for all integers less than or equal to a for any positive integer n. Now consider f(a+1) = an(a+1)n + …..+ a0, on expanding and using equation (3) we will get f(a+1) = q(a+1) + a0, therefore {f(a+1) – f(b)}/ (a+1) – b is an integer. This closes the induction and proves that (a – b) divides f(a) – f(b) for any integer a and an integer b lies between 0 and n.

Finally consider any distinct integers a and c, f(a) – f(c) can be written as {f(a) – f(b)} – { f(c) – f(b)}, where b lies between 0 and n, therefore f(a) – f(b) = K(a – b) and f(c) – f(b) = K(c – b), { from the last paragraph}. Therefore a – c divides f(a) – f(c).

Hence the proof.

Dear Beni,

I have highlighted a point in the proof, it is written in Caps lock.

It is not correct. Please try to bring strong arguments to your conclusions. What you say is that the polynomials are of degree , and this is again not true.

could you give me your solution?