### 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.

# Sums of Pairs

##### Age 11 to 16Challenge Level

Here's some sound and efficient algebra from Conor from Queen Elizabeth's Hospital :

We have three numbers, $x, y$ and $z$.

$x + y = 11 \quad (1)$
$y + z = 17 \quad (2)$
$z + x = 22 \quad (3)$

Adding $(1)$ and $(2)$ gives :

$x + 2y + z = 28$

Putting that another way :

$(x + z) + 2y = 28 \quad (4)$

Substituting $(3)$ into $(4)$ gives :

$22 + 2y = 28$

So $y = 3$.

Substituting $y = 3$ into $(1)$ gives $x = 8$ and substituting it into $(2)$ gives $z = 14$.

But Eden has had an excellent approach too :

$11 + 17 + 22$ uses every number twice and makes a total of $50$

So the total of the three numbers that we need to find is $25$.

If two of them make $11$ together the one not used must have been $14$.

Two together made $17$, the one not used this time must have been $8$.

And finally, a pair have a sum of $22$, so the other number is $3$.

The three numbers are $3, 8,$ and $14$

And Mark used some deductive reasoning combined with an exhaustive approach:

The smallest number must be between $1$ and $5$ to make a smallest sum of $11$.
If it is $1$ then to make $11$ the second number is $10$.
If the third number is $x$
Then $x+1= 17$
and $x+10 = 22$
This does not work.

I tried the smallest number as $2$, then the second number is $9$
Then $x+2= 17$ and $x+9 = 22$
This also does not work but the two values of $x$ are closer than last time so I think $3$ will work.

Trying the first number as $3$ and the second as $8$
Then $x+3= 17$ and $x+8 = 22$
This works.

The three numbers are $3, 8$ and $14$.