0
$\begingroup$

For the purpose of this question and the patterns pointed out, $0$ is not a perfect square (some think it is, others no).

The first perfect squares are $1$, $4$, $9$, $16$, $25$ and so on. The difference between two consecutive squares, each time, increases by $2$.

$$2^2 - 1^2 = 3$$ $$3^2 - 2^2 = 5$$ $$4^2 - 3^2 = 7$$

A pattern can also be established based on the perfect square's position.

Given a perfect square's position ($1$ is the first perfect square, $4$ is the second perfect square), the next perfect square can be produced by:

$$PerfectSquare + 2(PerfectSquarePosition) + 1$$

$1$ is the first perfect square. The next perfect squares ($4$) is $1 + 2(1) + 1$. Since $4$ is the second perfect square, the perfect square after $4$ is $4 + 2(2) + 1$. The perfect square after $9$ is $16$, which equals $9 + 2(3) + 1$.

Is there such a pattern for perfect cubes? I was unable to find a way to consistently relate $1$, $8$, $27$ and so one, both through the first pattern I pointed out (the difference between two consecutive cubes) and the second pattern (finding the next perfect square based on the previous perfect square and its position).

$\endgroup$
  • 1
    $\begingroup$ There's nothing wrong with considering $0$ a perfect square (and it's strange not to). Your pattern still holds; think of $0$ as the "zeroth" perfect square, one before the first. The pattern even continues backwards into the negative numbers: $$\ldots$$ $$(-1)^2 - (-2)^2 = -3$$ $$0^2 - (-1)^2 = -1$$ $$1^2 - 0^2 = 1$$ $$2^2 - 1^2 = 3$$ $$\ldots$$ $\endgroup$ – Théophile Jul 6 '16 at 21:32
1
$\begingroup$

$$(n+1)^3 - n^3 = 3n(n+1) +1$$

For example,

$$10^3 - 9^3 = 3\cdot 9 \cdot 10 + 1 = 271 \\ 1000 - 729 = 271 $$

That pattern looks a little like the pattern for squares...

$\endgroup$
1
$\begingroup$

Don't just take differences, take iterated differences!

The differences of the cubes are 7, 19, 37, 61, 91, ..., the differences of these numbers are 12, 18, 24, 30, ..., and the differences of these numbers are all 6. (This is still true if you allow 0 as perfect cube.)

$\endgroup$
  • $\begingroup$ Interesting... and with perfect 4ths the difference of all the differences is 36... the square of 6. Thanks for pointing it out :) $\endgroup$ – Mar Dev Jul 6 '16 at 21:35
  • $\begingroup$ No, the fourth differences of the fourth powers are 24, which is by no coincidence $4!$, just like the fourth derivative of $x^4$ $\endgroup$ – Christian Sievers Jul 6 '16 at 21:41
  • $\begingroup$ Unless I am missing something, 0 1 16 81 -> 1 15 65 -> 14 50 -> 36 $\endgroup$ – Mar Dev Jul 6 '16 at 21:53
  • $\begingroup$ 0 1 16 81 256 -> 1 15 65 175 -> 14 50 110 -> 36 60 -> 24 $\endgroup$ – Christian Sievers Jul 6 '16 at 21:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.