Divisiblity by 5

Number Theory Level 2

If we assume that n satisfies the equation:

$5 | n^{2} -n-2$

Then what can we say about this equation for n?

$n^{3}+n^{2}-2n$

There's a finite number of n that makes the equation not divisible by 5. There's a finite number of n that makes it divisible 5. There's not enough information to say any of the above. The equation will always be divisble by 5. The equation can never be divisible by 5.

1 solution

Mads Pasztor
Jul 15, 2018

We are given that $5 | n^{2}-n-2$ . We can factorize that to make it easier.

By factorization we get: $5 | (n+1)(n-2)$

Because 5 is a prime, we know that either $5 | n+1$ is true, or $5 | n-2$ is true.

Now let's take a look at our second equation: $n^{3} + n^{2} - 2$ . We can also factorize this.

By factorization we get: $n(n-1)(n+2)$

We can prove this can never be divisible by 5 by using contradiction.

Assume that $5 | n(n-1)(n+2)$

Again since 5 is a prime number this means that either $5|n$ is true, or $5|n-1$ is true, or $5|n+2$ is true.

Let's consider the numbers: $n-2, n-1, n, n+1, n+2$

Those are 5 consecutive numbers, which mean there is exactly 1 number in those 5 which is divisible by 5.

However our first equation says that either $n+1$ or $n-2$ is divisible by 5.

And our second equation, if true, tells us that either $n$ or $n-1$ or $n+2$ is divisble by 5.

This is, however, a contradiction since that means 2 numbers in $n-2, n-1, n, n+1, n+2$ is divisible by 5, which is not true.

So therefore we must conclude that our assumption was false and therefore $n(n-1)(n+2)$ or $n^{3} + n^{2} - 2n$ is never divisible by 5.

Divisiblity by 5

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