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Any non prime number that >= 4 can be written in the format: i^2 + i * k.

Condition: i>=2, k >= 0 (i, k are integers)

4 = 2^2 + 2*0 -> Good

6 = 2^2 + 2*1 -> Good

10 = 2^2 + 2*3 -> Good ...

Thanks!

LHA
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    Are $i, k$ integers? Are either of them allowed to be $0$. Notice that $i^2 + ik = i(i+k)$ and if $m=ab$ is not prime then $m = ab = a(a + (b-a)) = a^2 + a(b-a)$. – fleablood Mar 05 '21 at 22:57
  • Please, learn to use Mathjax to write your questions. – jjagmath Mar 05 '21 at 23:00

2 Answers2

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I observe $$ i^2 + ik = i(i+k) \text{.} $$

Let $n$ be a composite number, that is a non-prime number and let $i$ be the smallest prime dividing $n$. We know $2 \leq i$. Also, $n/i > 1$ and because $i$ is the smallest prime divisor of $n$, $n/i$ is an integer with $n/i > i$. So we have $$ 2 \leq i \leq \frac{n}{i} < n $$ Let $k = \frac{n}{i} - i$. By the above inequality, this $k \geq 0$. Also, \begin{align*} i^2 + ik &= i^2 + i\left(\frac{n}{i} - i \right) \\ &= i^2 + \left(n - i^2 \right) \\ &= n \text{,} \end{align*} as desired.

Eric Towers
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This isn't as deep as it looks.

$i^2 + ik = i(i + k)$.

And if you have any $m \le n$ then set $i = m$ and $k = n -m \ge 0$ and you are done.

If $J$ is not prime then $J = mn$ for some $1 < m\le n < J$. So if $i = m \ge 2$ and $k =n-m \ge 0$ then we have $i(i+k) = mn = K$.

That's all.

(although it might be strange to think that $J = mn = m^2 + m(n-m)$ but a little arithmetic shows that must be the case.)

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Okay..... just for schlits and giggees. Here's the hard way to do it:

Let $J = mn; m \le n$.

Then $m^2 \le mn = J \le n^2$

So $J = m^2 + W$ where $W = J - m^2$. But $J= mn$ so $W = mn -m^2 = m(n-m)$ and so $m|W$. Let $k = \frac Wm$.

Then $J= m^2 + W = m^2 + mk$.

fleablood
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