User:RandomUndergrad/Sandbox/Henry Ernest Dudeney/Puzzles and Curious Problems/109 - Perfect Squares/Solution

by : $109$

 * Perfect Squares

Solution
The smallest such set seems to be:


 * $10 \, 430, 3970, 2114, 386$

We have:

Proof
Dudeney says: "The general solution depends on the fact that every prime number of the form $4 m + 1$ is the sum of two squares. Readers will probably like to work out the solution in full."

The sum of all four numbers is square and can be expressed as a sum of two squares in three ways.

Hence we look at Pythagorean triples.

Following Dudeney's hint, we arrive at the smallest Pythagorean triples $\tuple {3, 4, 5}$ and $\tuple {5, 12, 13}$.

The lowest common multiple of $5$ and $13$ is $64$, and is itself a sum of two squares:
 * $65 = 8^2 + 1^2 = 7^2 + 4^2$

which generates the Pythagorean triples $\tuple {63, 16, 65}$ and $\tuple {33, 56, 65}$.

Thus, by multiplying out the Pythagorean triples we obtained, we arrive at the equation:
 * $65^2 = 63^2 + 16^2 = 39^2 + 52^2 = 25^2 + 60^2$

Now we attempt the solve the equations:

Subtracting the fourth equation from the sum of the rest, we obtain:

Note that if $y^2$ and $z^2$ are replaced with $\paren {65^2 - y^2}$ and $\paren {65^2 - z^2}$, we would have:

so we see that when $y, z$ are swapped, $a, b$ and $c, d$ would simply be swapped.

let $x = 63$.

Then by the above we only need to consider two cases:
 * $(x, y, z) = (63, 39, 25)$ or $(63, 39, 60)$

The first case gives the solutions
 * $a = 945, b = 3024, c = 576, d = -320$

The second case gives the solutions
 * $a = 4865/2, b = 3073/2, c = -1823/2, d = 2335/2$

which both contain negative numbers, and is not what we are looking for.

Check again for the inclusion of $\tuple {33, 56, 65}$. This should yield a positive noninteger solution.

Repeat the above for the hypotheuses $85 = 5\times 17$ and $125 = 5 \times 25$.

Then the next candidate should be $130$, and the solution would be the noninteger solution $\times 4$.