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Upsetting Pitagoras

Find the smallest integer solution to the equation 1/x^2 + 1/y^2 = 1/z^2

Rudolff's Problem

A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?

Diophantine N-tuples

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

Euler's Squares

Age 14 to 16
Challenge Level

Congratulations to Tony Cardell, age 14, State College Area High School, Pennsylvania, USA for this solution. Three of the numbers Euler has listed are 18530, 65570 and 45986. We want to find the fourth number that will complete his set so that any two added together form a perfect square. Therefore, we can set up the equations, where $p$, $q$ and $r$ are natural numbers, and $x$ is our fourth Euler number:

$\begin{eqnarray} 18530 + x &=p^2\\ 65570 + x &=q^2\\ 45986 + x &=r^2, \end{eqnarray}$

Now by subtraction, we have $q^2-r^2 = 19584.$ Following the hint given in the problem, we know $q \geq \sqrt{65570} \geq 257$ and $r \geq \sqrt{45986} \geq 215$\ so $q+r \geq 472$ and \[ q-r \leq {19584 \over 472} \leq 41.5 \] so $q-r$ is less than or equal to $41$.

Now since we have the factorization $(q - r)(q + r)$, we want to find possible values of $q - r$ in our range from $1$ to $41$. You can do this by computing the prime factorization of $19584$ which is $2^7 \times 3^2 \times 17$. This generates a table of small factors. Here are the ones under $41$: $1$, $2$, $3$, $4$, $6$, $8$, $9$, $12$, $16$, $18$, $24$, $32$, $34$, $36$. From these values we can find $q+r$ easily. Adding $q-r$ and $q+r$ yields $2q.$ We want as small $q$ as possible (we want to keep them around Euler's other numbers), so since $q-r$ and $q+r$ are inversely related, we should start the calculation of possible $q$'s with the factors closest to the squares: $36$, $34$, and $32$. Each of these yields $q$ values respectively of $290$, $305$, and $322$. Squaring these and subtracting $65570$ yields possible $x$ values. Respectively these are: $18530$, $27455$, $38114$. Now $18530$ is already on Euler's list, so we move on to the next one, $27455$. We find this fails when added to $18530$ (it does not form a perfect square in this case). Moving on to the next possible value we find: combinations of $18530$, $65570$, and $45986$ are given to work among themselves.

$\begin{eqnarray} 38114 + 18530 &=238^2\\ 38114 + 65570 &=322^2\\ 38114 + 45986 &=290^2. \end{eqnarray}$

Thus $38114$ is our answer!!