### Diophantine N-tuples

Can you explain why a sequence of operations always gives you perfect squares?

### DOTS Division

Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.

### Sixational

The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.

# Leftovers

##### Age 14 to 16 ShortChallenge Level

Answer: 9 different remainders (0, 1, 2, 3, 4, 5, 6, 7, 16)

$n$ $n^2$ $n+4$ remainder
1 1 5 1
2 4 6 4
3 9 7 2
4 16 8 0
5 25 9 7
6 36 10 6
7 49 11 5
...     ...
Not much pattern so far
...     ...
20 400 24 24$\times$15 = 240 + 120 = 360
24$\times$16 = 360 + 24 = 384
remainder is 16
100 10000 104 104$\times$100 = 10400
104$\times$96 = 10400 - 4016
= 10000 - 16
remainder is 16
101 10201 105 105$\times$100 = 10500
105$\times$97 = 10500 - 315
= 10200 - 16
remainder is 16

Remainder is 16 for the larger numbers tested
Had to multiply by $n-4$ for the larger values of $n$, try this algebraically: $(n-4)(n+4)=n^2-16$

So $n^2 = \underbrace{(n-4)(n+4)}_{\text{multiple of }n+4} + 16$

So the remainder is the same as the remainder when $16$ is divided by $n+4$
This will be $16$ if $n+4\gt16$ so if $n\gt12$.

Check values for $n=8$ to $n=12$ ($n$ up to $7$ shown above)
$n$ $n+4$ remainder when $16$ divided by $n+4$
8 12 4
9 13 3
10 14 2
11 15 1
12 16 0
So possible remainders are 0 - 7 and 16, 9 possibilites.

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.