Bob

]]>In the middle of the night I had a further thought and I now think the last lines above need this addition.

(n-a)(n-b)(n-c)(n-d)Q(n) = 0 + 13

**It is possible to choose 'n' and 'a' so that (n-a) = 13 (say) and 'b' so that (n-b) = 1 and 'c' so that (n-c) = -1 and Q so that Q(n) = -1, but, as a, b, c, d are said to be distinct, there is no other factor of 13 in the set of integers for (n-d) and therefore ....**

This is a contradiction as 13 cannot have these many (distinct) integer factors, so the assumption was false.

]]>P(x) = (x-a)(x-b)(x-c)(x-d)Q(x) for a, b, c, d all integers and Q is some polynomial of lower power than P.

Now assume 'to the contrary' that n exists,

ie P(x) = (x-n)R(x) + 13 where R is also a polynomial of lower order than P

Put x = n in both expressions

(n-a)(n-b)(n-c)(n-d)Q(n) = 0 + 13

This is a contradiction as 13 cannot have these many (distinct) integer factors, so the assumption was false.

Hope this helps,

Bob

]]>13 only has two factors - 1 and 13 - but here we've shown that any number produced using this polynomial will have at least 4.

(The exception being when you use a value of n that causes P(n) = 0, but P(n) isn't 13 in that case either)

If

is a polynomial with integer coefficients, and that this polynomial takes the value zero at four distinct integer values a, b, c and d.

Show that there exists no integer value n for which

.I don't know how to do it and my book doesn't go through it well enough. Can anyone help me?

Thanks

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