*The polynomial P(x) has the property that P(1), P(2), P(3), P(4), and P(5) are equal to 1, 2, 3, 4, 5 in some order. How many possibilities are there for the polynomial P, given that the degree of P is strictly less than 4?*

*(Duke Math Meet 2013 Tiebreaker round)*

**Discussion:**

Let

Suppose P(1), P(2), P(3), P(4) are p,q,r,s respectively. Then in matrix notation we may write:

Here p, q, r, s is a permutation of a selection from 1, 2, 3, 4, 5.

This implies:

Note that or

This implies

Next we work with cases:

**P(5) = 2 or 4 (even)** implies p is even (hence 4 or 2 respectively). Since otherwise LHS will odd.

Hence . or implies .

Therefore is divisible by 3. This is possible only when q+s = 1+ 5 or 5 +1 and r = 3.

**Hence P(5) =2 implies (p,q,r,s) = (4, 1, 3, 5) or (4, 5, 3, 1); P(5) = 4 implies (p,q,r,s) = (2, 1, 3, 5) or (2, 5, 3, 1);**

**P(5) = 1, 3 or 5** implies p is odd (to maintain parity)

Then = 4 (1+3 or 3+1) , 6 (1+5 or 5+1), 8 (3+5 or 5+3)

For each of these sub cases there are 4 ‘solutions’ of (p,q,r,s). (found by similar computations).

*Example: if then . Clearly q+s is divisible by 3. But that is possible iff q+s = 2+4. This implies r =3.*

**Hence number of valid polynomials are 16**

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