ISI2019-MMA-12

Question

Given a positive integer $m$, we define $f(m)$ as the highest power of $2$ that divides $m$. If $n$ is a prime number greater than $3$, then

• A. $f(n^3-1) = f(n-1)$
• B. $f(n^3-1) = f(n-1) +1$
• C. $f(n^3-1) = 2f(n-1)$
• D. None of the above is necessarily true

Comments

Plz enlarge the size.not readable

Answer 1

We can write $(n^3 - 1) = n^3 - 1^3$

Now, applying the formula $$a^3 - b^3 = (a – b)(a^2 + ab + b^2)$$

$$\implies n^3 - 1 = (n - 1)(n^2 + n + 1)$$

We observe that $(n^3 - 1)$ is divisible by $(n - 1)$.

Since $(n^3 - 1)$ is divisible by $(n - 1)$, so, if some $2^k$ divides $(n - 1)$, then that same $2^k$ will divide $(n^3 - 1)$ as $(n - 1)$ is a factor of $(n^3 - 1)$. We don't take into account $(n^2 + n + 1)$ as it is always odd, i.e., for both even and odd $n$. So, $(n^2 + n + 1)$ is not divisible by $2$.

Therefore, $f(n^3 - 1) = f(n - 1)$.

This is true for all $n > 2$.

For $n = 2$, $$(n^3 - 1) = (2^3 - 1) = 8 - 1 = 7$$ which is the only prime number of the form $(n^3 - 1)$. Since in question, they have clearly written that $n > 3$, the answer is $$f(n^3 - 1) = f(n - 1)$$

i.e., Option $(A)$.

Comments

what about $n^2+n+1$ . It might also be even. Your argument establishes the fact that  $f(n^3-1)\geq f(n-1)$ . Not equality.

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If $n$ is even, $n^2 + n + 1$ will be an odd number. And if $n$ is odd, then too $n^2 + n + 1$ will be odd. :)

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$n^2 + n + 1$ will not be even.

Answer 2

$n^3 -1=(n-1)(n^2+n+1)$
$n>3$ is a prime number
$\implies n^3, n^2, n $ are odd numbers

$\implies n^3-1, n-1$ are even numbers, and $n^2+n+1$ is odd number because sum of three odd numbers is odd.

That means $n^2+n+1$ has no hidden powers of 2 and $\implies n^3-1, n-1$ have equal number of factors of 2.
$\implies f(n-1)= f(n^3 -1)$ where $f$ is a function which outputs the highest power of $2$ (given in question).

So $A$ is correct.

Answer 3

n^3 – 1 = (n – 1)(n^2 + n + 1)

also, n^2 + n + 1 = odd, as n^2 is odd, n is odd ad 1 is odd.

Hence, if f(n – 1) = x, then f(n^3 – 1) is also x, as only (n – 1) part contains 2 as factor and the other part is odd.

So, f(n – 1) = f(n^3 – 1)