Challenge Level

**1. What is $(a^2 + d^2) - (b^2 + c^2)$ equal to? Show that it always takes this value.**

Yuk-Chiu from Harrow School, Ryan from Stowe School and Nayanika from the Tiffin Girls' School, all in the UK used algebra, starting from the first number. This is Nayanika's work:

Nishad from Thomas Estley Community College in the UK also used algebra, but started from the second number:

Firstly lets write $a, b, c$ and $d$ as:

$$a=x-1 \\ b=x \\ c=x+1 \\ d=x+2$$

We can say that

$$\begin{split}(a^2+d^2)-(b^2+c^2) &= ((x-1)^2 + (x+2)^2) - (x^2 + (x+1)^2)\\ &=(2x^2 + 2x +5) - (2x^2 +2x +1) \\ &= 4\end{split}$$ which is independent of $x$

**2. Why is $(a^2 + b^2 + c^2 + d^2) - (1+2+3)$ always divisible by $4$?
Why is it always divisible by $8$?**

Yuk-Chiu and Ryan used algebra and factorisation to show why this is true. Yuk-Chiu's work:

Ryan explained the last step in a bit more detail:

We have to turn our attention to the second half of the equation: $(x+1)(x+2).$ We know that if two numbers multiply together, as long as one of them is an even number, the product is going to be an even number as well. Here, for $x+1)(x+2),$ whatever the number is, we will always have an even number and an odd number multiply together. So we can always get an even product. And we already know that the whole equation equals times 4, so 2×4=8, and we finally find out that the whole equation is divisible by 8.

Nishad and Nayanika used a slightly different argument. This is Nayanika's work:

Because Nishad used $x$ for the second number, not the first, Nishad's proof is simpler:

$$\begin{split}(a^2+b^2 + c^2+d^2) - (1+2+3)&= (x-1)^2 + x^2 + (x+1)^2 + (x+2)^2 -6\\&=4x^2 + 4x = 4(x^2 + x)\end{split}$$

so [it] is divisible by $4$

Now looking at the bit in the brackets, namely $x^2+x$ it is one of the two following cases:

odd + odd = even

even + even = even

So $x^2+x$ is always divisible by $2$ making the whole number divisible by $8$

Nishad also explored the generalised problem by conjecturing that 'The sum of the square of $n$ consecutive integers [subtract the first $n-1$ integers] is divisible by $n$'. Click to see Nishad's extension work.

Conjecture: $$\sum_{r=1}^{n} (a_r)^2 - \sum_{r=1}^{n-1} r$$ is divisible by $n$ $\{

a_r$ are consecutive integers $\}$.

*Nishad's conclusions are correct, but there are some mistakes in Nishad's algebra. Can you correct them?*

I will split this into 2 cases, $n=2k-1$ and $n=2k$ $\{ k \in \mathbb{Z}^+\}$ (*so $k$ is a positive integer*)

Case 1: $n=2k-1$

$$\sum_{r=1}^{2k-1} (x-k+r)^2 - \sum_{r=1}^{2k-2} r =

\sum_{r=1}^{2k-1} \left[ x^2 + 2(r-k)x +(r-k)^2 \right] - \sum_{r=1}^{2k-2}

r\\=(2k-1)x^2 + [2k(2k-1) -2k(2k-1)]x + \frac{(k-1)k(2k-1)}{3} -

\frac{(2k-2)(2k-1)}{2}\\=(2k-1)x^2 + \frac{k(2k-1)(k-1)}{3}-(k-1)(2k-1)$$

Now we can divide through by $(2k-1)$ we can check whether the above expression is divisible by $(2k-1)$.

$$x^2 + \frac{k(k-1)}{3} - (k-1)$$

The above is an integer when either $k$ is a multiple of 3 or 1 more than a multiple of 3 but NOT when k is 2 more than a multiple of 3 (which does disprove the conjecture for all $n$)

Case 2: $n=2k$

$$\sum_{r=1}^{2k} (x-k+r)^2 - \sum_{r=1}^{2k-1} r$$

(We can use the answer to the last case to help evaluate this as:)

$$=2kx^2 + 2kx + \frac{k(2k-1)(k-1)}{3} + k^2 -(k-1)(2k-1)+(2k-1)\\=2kx^2 + 2kx + \frac{k(2k^2+1)}{3} -(2k-1)k$$

Now we can divide through by $2k$ we can check whether the above expression is divisible by $2k$.

$$x^2 + x + \frac{2k^2+1}{6} -\frac{2k-1}{2} = x^2 + x +\frac{(k-2)(k-1)}{3}$$

The above is an integer when either $k$ is 1 more than a multiple of 3 or 2 more than a multiple of 3 but NOT when k is a multiple of 3.

In conclusion the conjecture is true when $n$ is 1 more than a multiple of 6, 2 more than a multiple of 6, 4 more than a multiple of 6 and 5 more than a multiple of 6 but this is equivalent to saying $n$ is not a multiple of 3 so:

$$n \neq 3k \iff n\mid \left( \sum_{r=1}^{n} (a_r)^2 - \sum_{r=1}^{n-1} r\right) \{ k \in \mathbb{Z}^+ \}$$

a_r$ are consecutive integers $\}$.

I will split this into 2 cases, $n=2k-1$ and $n=2k$ $\{ k \in \mathbb{Z}^+\}$ (

Case 1: $n=2k-1$

$$\sum_{r=1}^{2k-1} (x-k+r)^2 - \sum_{r=1}^{2k-2} r =

\sum_{r=1}^{2k-1} \left[ x^2 + 2(r-k)x +(r-k)^2 \right] - \sum_{r=1}^{2k-2}

r\\=(2k-1)x^2 + [2k(2k-1) -2k(2k-1)]x + \frac{(k-1)k(2k-1)}{3} -

\frac{(2k-2)(2k-1)}{2}\\=(2k-1)x^2 + \frac{k(2k-1)(k-1)}{3}-(k-1)(2k-1)$$

Now we can divide through by $(2k-1)$ we can check whether the above expression is divisible by $(2k-1)$.

$$x^2 + \frac{k(k-1)}{3} - (k-1)$$

The above is an integer when either $k$ is a multiple of 3 or 1 more than a multiple of 3 but NOT when k is 2 more than a multiple of 3 (which does disprove the conjecture for all $n$)

Case 2: $n=2k$

$$\sum_{r=1}^{2k} (x-k+r)^2 - \sum_{r=1}^{2k-1} r$$

(We can use the answer to the last case to help evaluate this as:)

$$=2kx^2 + 2kx + \frac{k(2k-1)(k-1)}{3} + k^2 -(k-1)(2k-1)+(2k-1)\\=2kx^2 + 2kx + \frac{k(2k^2+1)}{3} -(2k-1)k$$

Now we can divide through by $2k$ we can check whether the above expression is divisible by $2k$.

$$x^2 + x + \frac{2k^2+1}{6} -\frac{2k-1}{2} = x^2 + x +\frac{(k-2)(k-1)}{3}$$

The above is an integer when either $k$ is 1 more than a multiple of 3 or 2 more than a multiple of 3 but NOT when k is a multiple of 3.

In conclusion the conjecture is true when $n$ is 1 more than a multiple of 6, 2 more than a multiple of 6, 4 more than a multiple of 6 and 5 more than a multiple of 6 but this is equivalent to saying $n$ is not a multiple of 3 so:

$$n \neq 3k \iff n\mid \left( \sum_{r=1}^{n} (a_r)^2 - \sum_{r=1}^{n-1} r\right) \{ k \in \mathbb{Z}^+ \}$$

**3. Why is $abcd$ divisible by $24$?**

Ryan explained why using the factors of $24$:

Four consecutive number multiply together, how is it divisible by $24$? We know if a number is divisible by $x,$ it must be divisible by any of the factors of $x.$

So think about $24,$ we will know that $24=2\times2\times2\times3.$ First, we know all the even numbers are divisible by $2,$ it means every two consecutive numbers, there is a number that is divisible by $2,$ there are $4$ consecutive number in total, so there should be two even numbers which are divisible by $2.$ So we solve the $2$ here, what’s left is $2\times3\times3.$ For being divisible by $3,$ we know that $3\times1=3,$ $3\times2=6,$ $3\times3=9,$ …, so every three consecutive numbers, there is a number which is divisible by $3.$ And we’ve got four consecutive numbers here! So there must be a number in it which is divisible by $3.$ So now we solve $3$ as well, what’s left is $2\times2.$ Until now, many of you may already know what’s going on and what to do here. $2\times2$ is $4.$ And every $4$ consecutive numbers there is a number that is divisible by $4.$ And we have exactly $4$ consecutive numbers here, so there must be a number which is divisible by $4$ as well. Now we know that in these four consecutive numbers, we will be able to find all $2, 3,$ and $4$ in the factors of these four numbers. So $abcd$ will be always divisible by 24.

Yuk-Chiu and Nishad explained why using the factors of the consecutive numbers. This is Yuk-Chiu's work:

**4. Explore $\sqrt{abcd + 1}$.**

Ryan's exploration included algebra and trials:

For the forth question, I started with algebra as usual. Let $a=x, b=x+1, c=x+2, d=x+3$ and put it into $\sqrt{abcd+1},$ and we get $\sqrt{x(x+1)(x+2)(x+3)+1}.$

We need to simplify this [expression] first.

$\sqrt{x(x+1)(x+2)(x+3)+1} = \sqrt{x^4+6x^3+11x^2+6x+1}$

Here we have to find a way to factorize the [expression]. Because we have $x^4$ so we will know there must be two $x^2$ multiply together, and because the only constant is $1$, so we know it can only be $1\times1.$ And it gives us this equation $(x^2+ax+1)(x^2+bx+1).$ Now the only thing we have to do is to find what’s $a$ and $b.$ And it’s not hard to try it a few times on paper to get the answer $a=b=3.$

So the [expression] will be: $$\sqrt{\left(x^2+3x+1\right)^2} = |x^2+3x+1|$$ Because here we have a square and a root, the square and root of a number is its absolute value (*since a positive number and its negation have the same square, but square rooting returns only the positive number*).

Now, we totally get rid of the root, so it shows us that $abcd+1$ is always a perfect root. And because $x$ is an integer, $x^2+3x+1$ will be an integer as well. The solution of $\sqrt{abcd+1}$ must be an integer.

Is that the end of it? Yes, but not necessarily. Let’s go back to the first equation, if we bring some constants into the equation, it will give us something like this:

We will then find a pattern from it, the solution of $\sqrt{abcd+1}$ is always $ad+1,$ but this is only the pattern we find, we still need to prove its generality. Let $a=x, b=x+1, c=x+2, d=x+3,$ and bring it into $ad+1:$

$\begin{split}ad+1 &= x(x+3)+1\\&=x^2+3x+1\end{split}$

$\sqrt{abcd+1} = |x^2+3x+1|$ and $ad+1 = x^2+3x+1$

$\therefore \sqrt{abcd+1} = |ad+1|$

Nishad factorised the expression in a different way, which led to an investigation involving the golden ratio:

We can say that $$\sqrt{abcd+1} = \sqrt{(x-1)x(x+1)(x+2) + 1}$$

$$=\sqrt{x^4 + 2x^3 -x^2 -2x + 1} = \sqrt{(x^2 +x-1)^2} = \sqrt{(x+\phi)^2

(x-(\phi - 1))^2}\\=(x+\phi)(x-(\phi - 1))$$

where $\phi$ is the golden ratio

*If you want to see where this comes from, do the first part of this problem. You'll derive $\phi$ via a quadratic equation whose solution is $\phi$.*

*If you let $b=1$ instead of $a,$ then the quadratic equation in $b$ has a solution equal to $\frac1\phi,$ which is equal to $\phi-1.$ You will then recognise this in the expression which Nishad factorses above.*

Side note: If we had used the consecutive integers as $(x-2),(x-1),x,(x+1)$ then we would get that

$$\sqrt{abcd+1} = \sqrt{(x^2 -x-1)^2} = (x-\phi) (x+(\phi -1))$$

and in fact any algebraic set of 4 consecutive numbers just translates the graph left or right.

The way I recognized the factorization of the polynomial is by considering the coefficients of the 'golden polynomials.' The 'golden polynomials' are of the form: $$(x^2 +x -1)^n$$

or

$$(x^2 -x -1)^n$$

After seeing this I questioned whether multiplying a different amount of consecutive numbers and adding 1 would still give 'golden polynomials' but unfortunately this was not true (however 2 consecutive numbers multiplied together then subtracted by 1 does give a 'golden polynomial').

Yuk-Chiu's work led to an interesting investigation about prime numbers:

Yuk-Chiu has shown that *all prime numbers greater than 3 can be expressed in the form $\sqrt{24n+1}$ for some value of $n$*, and also that *the product **four consecutive integers $abcd$ can be expressed in the form $24m.$*

Yuk-Chiu has not shown (or claimed) that *$\sqrt{24n+1}$ is always prime*, that *every multiple of 24 can be written as the product of four consecutive integers*, that *$\sqrt{abcd+1}$ is always a prime number*, or that every prime number can be written in the form $\sqrt{abcd+1}$. Which of those statements are true?