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

Now by subtraction, we have

q^{2} - r^{2} = 19584 .

Following the hint given in the problem, we know

q ≧ \sqrt{6} 5570 ≧ 257

and

r ≧ \sqrt{4} 5986 ≧ 215

q + r ≧ 472

and

q - r ≦ \frac{19584}{472} ≦ 41.5

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 $2 q .$

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.

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

Thus 38114 is our answer!!