I recently ran into the following interesting problem: Can you find distinct integers such that

$$

\begin{eqnarray*}

a^{2}+b^{2}+c^{2}+d^{2} & = & abcd?

\end{eqnarray*}

$$

I like this problem because a friend recently told me there are no interesting problems accessible to high school students, and that all the interesting ones require advanced math degrees. This is one of those problems that crushes that argument. Not only is it accessible to high school students, but it is a fun and interesting problem to investigate. I follow with only a small insight in order to leave those curious enough with plentyto discover, and investigate. . . Have fun!

Here we will investigate one possible approach.

Taking the original equation, and completing the square we get,

$$

\begin{eqnarray*}

a^{2}+b^{2}+c^{2}+d^{2} & = & abcd\\

a^{2}-2abcd+(bcd)^{2}+b^{2}+c^{2}+d^{2} & = & abcd-2abcd+(bcd)^{2}\\

(a-bcd)^{2}+b^{2}+c^{2}+d^{2} & = & -abcd+(bcd)^{2}\\

& = & (bcd-a)bcd\\

\Rightarrow(bcd-a)^{2}+b^{2}+c^{2}+d^{2} & = & (bcd-a)bcd

\end{eqnarray*}

$$

and

$$

a'=bcd-a

$$

or equivalently,

$$

a_{n}=bcd-a_{n-1}

$$

We chose to complete the square using 'a', but what about b, c, or

d, after all addition and multiplication are commutative. Well it

similarly follows for those. Rewriting the original equation we get,

$$

x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=x_{1}x_{2}x_{3}x_{4}

$$

thus,

$$

x_{i}^{'}=\prod_{_{\begin{array}{c}

j=1\\

j\neq i

\end{array}}}^{4}x_{j}-x_{i}

$$

We see that,

$$

\begin{eqnarray*}

a_{1} & = & bcd-a_{0}\\

a_{2} & = & bcd-bcd+a_{0}\\

& = & a_{0}\\

a_{3} & = & bcd-a_{0}

\end{eqnarray*}

$$

which has a period of two. Thus, we can't just endlessly continue

iterating the same respective term, but from 'part II' we can see

we can change it up. One way could be,

$$

(a,b,c,d)\rightarrow(a',b,c,d)\rightarrow(a',b',c,d)\rightarrow(a',b',c',d)\rightarrow(a',b',c',d')\rightarrow(a'',b',c',d')

$$

and so on, but this blows up rather quickly.

Another way could be to evaluate,

$$

x_{n}x_{n-1}x_{n-2}-x_{n-3}

$$

and assigning it to,

$$

x_{n+1}

$$

consequently generating the sequence,

$$

\{x_{1},x_{2},x_{3},x_{4},...,x_{n}=x_{n}x_{n-1}x_{n-2}-x_{n-3},...\}

$$

of which any four consecutive terms in the sequence satisfies the

original equation.

For example, using the trivial solution $\{2,2,2,2\}$ yields,

$$

2

$$

$$

2

$$

$$

2

$$

$$

2

$$

$$

6

$$

$$

22

$$

$$

262

$$

$$

34582

$$

$$

199330642

$$

$$

1806032092550706

$$

$$

12449434806576800059248920402

$$

$$

4481765860945171681908664776799089162954814190172722

$$

$$

\vdots

$$

and we can go on indefinitely.

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