### 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.

# Pair Products

##### Age 14 to 16Challenge Level

Kira, from the UK, started off by doing some calculations:

Say your numbers were 4, 5, 6, 7. When you times the outer numbers 4 and 7 it will equal 28, and when you times the two middle numbers 5 and 6 it will equal 30.

Let's try it again on four new consecutive numbers, such as 1, 2, 3, 4: when you times the outer numbers 1 and 4 it equals 4, and when you times the two middle numbers 2 and 3 it equals 6.

Conclusion: the product of the outer numbers will always be 2 less than the product of the numbers in the middle.

The class at Elgin Academy had the clever idea of testing this for some negative numbers too!

Tom drew rectangles, like in Alison's explanation:

If we overlap the two rectangles we cut a piece off at the bottom and create a new piece at the right hand side. The difference between the two small rectangles is always two. This works for all numbers, not just $9, 10, 11, 12$.

Sam, from Park Grove Primary, gave the following explanation:

If you choose the first number of the set to be n, then the outside multiplication will be n(n+3), which simplifies to $n^2+3n$.
The inside multiplication will be (n+1)(n+2), which simplifies to $n^2+3n+2$.
If you look back at the outside multiplication, you can see that it is 2 less than the inside multiplication, proving the answer.

Great! Nathan also gave a convincing explanation:

Call the number halfway in between the middle two x.
This is the second number in our list plus 0.5, or the third number minus 0.5, so if we multiply the second and third numbers together we will get $x^2-0.25$.
But x is also the first number plus 1.5, or the fourth number minus 1.5, so if we multiply them together we will get $x^2-2.25$.

Nice - here Nathan is making use of the identity: $(x-a)(x+a) = x^2 - a^2$. Well done!

Keone, from Sage Ridge School in Reno, started with four consecutive whole numbers:

The product of the first and last numbers is always $2$ less than the product of the middle two numbers.

Explanation: Suppose the first number is $x$. Then the second number is $x+1$, the third is $x+2$, and the fourth is $x+3$.

So the product of the first and fourth numbers is $x(x+3) = x^2 + 3x$.

Also, the product of the second and third numbers is $(x+1)(x+2) = x^2 + 3x + 2$.

So $(x+1)(x+2) = x(x+3) + 2$ for any chosen value of $x$.

When he considered five consecutive whole numbers he found that:

The product of the first and last numbers is always $3$ less than the product of the second and fourth numbers.

Explanation: Again, let the first number be $x$; then the second number is $x+1$, the third is $x+2$, the fourth is $x+3$, and the fifth is $x+4$.

So the product of the first and last numbers is $x(x+4) = x^2 + 4x$.

Also, the product of the second and fourth numbers is $(x+1)(x+3) = x^2 + 4x + 3$.

So $(x+1)(x+3) = x(x+4) + 3$ for any chosen value of $x$.

And with $n$ consecutive whole numbers he found that:

The product of the first and last numbers is always $n-2$ less than the product of the second and penultimate numbers.

Let us consider the general case where there are n consecutive whole numbers.

As before, let the first number be $x$; then the second number is x+1, the third is x+2, and so on.

The last number (the $n$th number) will be $x+n-1$. Thus, the penultimate (second-to-last) number will be $x+n-2$.

So the product of the first number and the last number will be $x(x+n-1) = x^2 + nx - x$

The product of the second and the penultimate numbers will be $(x+1)(x+n-2) = x^2 + nx - 2x + x + n - 2 = x^2 + nx - x + n - 2$

So $(x+1)(x+n-2) = x(x+n-1) + n - 2$;

that is, the product of the second and penultimate numbers will always exceed the product of the first and last numbers by exactly $n - 2$.

For example, if we take 6 numbers, the product of the $2$nd and $5$th numbers will be $4$ more than the product of the $1$st and last numbers.

We received similar findings from Aisling, from Grand Avenue Primary School:

The product of the first and last numbers of a series of consecutive whole numbers, is the same as the product of the second and penultimate numbers of that series, minus the number of numbers separating the first and last numbers.

Natasha, from the European School, generalised her findings in a similar way and then went on to check her conclusion:

We can generalise this problem by substituting particular numbers with letters.

Let the first number be $a$.

If there are $n$ numbers, where a is the first, then the last number is $(a+n-1)$.

The second number is $(a+1)$, the penultimate number $(a+n-2)$.

By multiplying the first and last numbers together, we get $a(a+n-1) = a^2 +an-a$

Multiplication of the second and penultimate numbers gives

$(a+1)(a+n-2) = a^2+an-a+n-2$

The difference therefore in the product of the first and last numbers and the product of the second and penultimate numbers is always $n-2$.

For added confirmation we can take a random example:

Consider the numbers $56, 57, 58, 59, 60, 61, 62$

where $n = 7$ and $a = 56$

Our general formula tells us that the difference in the product pairs should be $n-2$ (i.e. $5$).

When we do the calculation, we get

$$57 \times61 - 56 \times62 = 3477 - 3472 = 5$$

This result corresponds with the general one established above.

Theo from Dulwich College generalised this problem further:

How would one go about finding the difference between the product of the first and last numbers and the product of the second and penultimate numbers, where each number is not consecutive but increases by possibly 2, 3 or 4? Could we even find a general rule for any sequence that increases by a constant? Turns out we can!

Let's take the sequence 11, 15, 19, 23, 27, 31

15 Ã— 27 - 11 Ã— 31 = 405$-$341 = 64

But now let's try and use this sequence to make a general rule:

Let us call '$a$' the first number in our sequence, '$n$' the number of terms in the sequence and $c$ the constant that we increase by between each two terms.

Therefore, in the example 11, 15, 19, 23, 27, 31;

$a$(first number)=11;    $n$(number of terms)=6;    $c$(constant by which we increase by each time)=4

In this example we can notice that the last term was 5 times $c$ (which in that case was 4) more than the first term:

31 - 5Ã—4 = 11

The penultimate term was 4 times $c$ (which in that case was 4) more than the first term:

27 - 4Ã—4 = 11

Remember that in that example $n$ is 6 so to get the last term, you have to add $(n-1)c$ and for the penultimate term, you have to add $(n-2)c.$

We can actually make a sequence of numbers with what we have just observed:

$a, a+c, ”¦ , a+(n-2)c, a+(n-1)c$

The product of the 1st and last term will be:

$\begin{split} a(a+(n-1)c)&= a(a+nc-c) \\ &=a^2+ anc -ca \\ &=a^2+(n-1)ca\end{split}$

The product of the 2nd and penultimate term will be:

$(a+c)(a+(n-2)c)$

$=(a+c)(a+nc-2c)$

$=a^2+ anc-2ca+ca+nc^2-2c^2$

$=a^2 + (n-2+1)ca + (n-2)c^2$

$=a^2 +(n-1)ca+ (n-2)c^2$

So, in short,

the product of the 1st and last term will be $a^2+(n-1)ca$

the product of the 2nd and penultimate term will be $a^2 +(n-1)ca+ (n-2)c^2$

Therefore, the product of the second and penultimate numbers will always be greater than the product of the first and last numbers by exactly $(n-2)c^2$.

We can see if this works with our initial example:

11, 15, 19, 23, 27, 31;

$n$(number of terms)=6;    $c$(constant by which we increase by each time)=4

(6-2)42= 64. This is indeed what we got at the start!

It is interesting to notice that $a$ (the first term) is irrelevant to the final answer.

This is also beautiful as it works with all rational and irrational numbers as well as imaginary and complex numbers:

$2+4i, 3+7i, 4+10i, 5 + 13i$

$n=4$ and $c=1+3i$

$(4-2)(1+3i)^2=2(1+3i+3i-9)=2(6i-8)= 12i -16$

$\begin{split}(3+7i)(4+10i)-(2+4i)(5+13i)&=(12+30i+28i-70)-(10+26i+20i-52)\\& = (58i-58)-(46i-42) = 12i - 16\end{split}$

Sure enough, the product of the 1st and last number is $12i-16$ greater than the product of the 2nd and penultimate numbers.