SOLUTION: Show that there is no cubic polynomial p(x) with integer coefficients such that p(1)=4 and p(3)=5.

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p(x) = ax³ + bx² + cx + d
p(1) = a + b + c + d = 4
p(3) = 27a + 9b + 3c + d = 5
p(3) - p(1) gives,
26a + 8b + 2c = 1
2(13a + 4b + c) = 1
The coefficents are all integers. So the bracketed term is always an itegral value. Therefore, no matter what the value of the bracketed term is, the lhs will always be even! (since the bracketed term is multiplied by 2)
But the rhs is odd!!
So it can never happen.