1980_Question_2

So my set was S = (33, 34, 35, 36, 37, 38, 39)
Here’s my way to the solution:
Let’s consider the set S=(q, q+1, q+2 ,\ldots ,q+6)
Part a
I will ignore this for now, as I will construct a polynomial with integer coefficients

Part b
P(n)=n.
This means that P(n)-n=Q(x), where Q(x) has factors, let’s say q, q+1, q+4, q+5, q+6.
Then we can write Q(x)=a(x-q)(x-q-1)(x-q-4)(x-q-5)(x-q-6).
Now I’ll just take a=1 for simplicity.

Then P(x)=(x-q)(x-q-1)(x-q-4)(x-q-5)(x-q-6)+x.
Now we go to part c!

Part c
Now we have to solve P(x)=(x-q)(x-q-1)(x-q-4)(x-q-5)(x-q-6)+x for some x in our set.
Let’s just say the one member that works is q+3.

Substituting in we get P(q+3)=-36+(q+3)=0.

Thus q+3=36 and q=33.

But our solution is not complete yet. We have to prove that q+2=35 is not a root of this polynomial.
This is pretty simple (but necessary) as direct substitution shows that it doesn’t work.

Now (a) is satisfied by default since all our roots are integers and a was set to be 1.

QED.

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