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Never Divides By 5
Problem
Given that $x$ is a positive integer prove that $f(x) = x^2 + x + 1$ will never divide by 5.
Solution
When $x$ is divided by 5 the possible remainders are 0, 1, 2, 3, and 4. The respective remainders of $x^2$ will be $0, 1, 4, 9 \equiv 4$, and $16 \equiv 1$. This can be seen more clearly in a table.
| $x$ | $x^2$ | $x^2 + x + 1$ |
|---|---|---|
| 0 | 0 | $0 + 0 + 1 = 1$ |
| 1 | 1 | $1 + 1 + 1 = 3$ |
| 2 | 4 | $2 + 4 + 1 = 7 \equiv 2$ |
| 3 | 4 | $3 + 4 + 1 = 8 \equiv 3$ |
| 4 | 1 | $4 + 1 + 1 = 6 \equiv 1$ |
Hence we show that $f(x)$ will never divide by 5.
Show that the result holds even if $x$ is a negative integer.
Investigate which other values of $n$ that will never divide into $f(x)$.
Problem ID: 318 (07 Apr 2007) Difficulty: 2 Star
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