### Be Reasonable

Prove that sqrt2, sqrt3 and sqrt5 cannot be terms of ANY arithmetic progression.

In y = ax +b when are a, -b/a, b in arithmetic progression. The polynomial y = ax^2 + bx + c has roots r1 and r2. Can a, r1, b, r2 and c be in arithmetic progression?

### Summats Clear

Find the sum, f(n), of the first n terms of the sequence: 0, 1, 1, 2, 2, 3, 3........p, p, p +1, p + 1,..... Prove that f(a + b) - f(a - b) = ab.

# Polite Numbers

##### Stage: 5 Challenge Level:

Sugam made a conjecture that the number of ways a number can be written as a polite number is one less than the number of odd factors and gave the following example:

There are four odd factors of $15 (1,3,5,15)$
$15=7+8$
$15=4+5+6$
$15=1+2+3+4+5$
So the number of ways of expressing $15$ as sum of two or more than two consecutive natural numbers is $4-1=3$

Yike looked at odd numbers, numbers of the form $4m+2$, and numbers of the form $pm$ where $p$ is an odd prime.

$n$ odd: $n$ can be written as $m + (m+1)$, where $m = \frac{n-1}{2}$

\eqalign {m + (m+1) &= \frac{n-1}{2} + (\frac{n-1}{2}+1)\cr &= \frac{n-1+n-1+2}{2}\cr &=\frac{2n}{2}\cr &=n \cr}
For example, $13 = 6 + 7$, $19 = 9 + 10$.

$n = 4m + 2$: $n$ can be written as $$(\frac{n-2}{4}-1) + (\frac{n-2}{4}) +(\frac{n-2}{4}+1)+(\frac{n-2}{4}+2)$$
For example, $18 = 3 + 4 + 5 + 6$, $34 = 7 + 8 + 9 + 10$.

$n = pm$ where $p> 2$ and prime:

$$n= \sum _{i=-\frac{p-1}{2}}^\frac{p-1}{2}m-i$$

Note that some terms in this sum may be negative for certain values of $m$ and $p$, but they will be cancelled out by the corresponding positive terms.

For example, $544 = 17 \times 32$
\eqalign{544 &= \sum_{i=-8}^{8} 32-i \cr &= 40+39+38+37+36+...+27+26+25+24}

$424=53 \times 8$
\eqalign{424 &= \sum_{i=-26}^{26}8-i \cr &= 34+33+32+...+2+1+0+-1+-2+...+-16+-17+-18 \cr &= 34+33+32+31+30+...+22+21+20+19}

Several people commented that all numbers are polite except powers of 2. The following diagrams can be used to show why powers of 2 are not polite.

This diagram shows what happens when you add together an odd number of consecutive numbers. The numbers either side of $n$ can pair off, giving $2n$ each time, so we end up with an odd multiple of $n$.

This diagram shows what happens when you add together an even number of consecutive numbers. This time, the numbers pair off to give $n+n+1=2n+1$ each time, so we end up with a multiple of an odd number, $2n+1$

Thus all polite numbers must have at least one odd factor, so no power of $2$ can be polite.