Henry Ernest Dudeney/Puzzles and Curious Problems/109 - Perfect Squares/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $109$

Perfect Squares
Find $4$ numbers such that the sum of every two and the sum of all four may be perfect squares.


Solution

The smallest such set seems to be:

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

We have:

\(\ds 10 \, 430 + 3970 + 2114 + 386\) \(=\) \(\ds 16 \, 900\) \(\ds = 130^2\)
\(\ds 10 \, 430 + 3970\) \(=\) \(\ds 14 \, 400\) \(\ds = 120^2\)
\(\ds 10 \, 430 + 2114\) \(=\) \(\ds 12 \, 544\) \(\ds = 112^2\)
\(\ds 10 \, 430 + 386\) \(=\) \(\ds 10 \, 816\) \(\ds = 104^2\)
\(\ds 3970 + 2114\) \(=\) \(\ds 6084\) \(\ds = 78^2\)
\(\ds 3970 + 386\) \(=\) \(\ds 4356\) \(\ds = 66^2\)
\(\ds 2114 + 386\) \(=\) \(\ds 2500\) \(\ds = 50^2\)


Proof


This theorem requires a proof.
In particular: 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."
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Sources

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