# Solution

(related to Problem: A Problem in Squares)

The sides of the three boards measure $31$ in., $41$ in., and $49$ in. The common difference of area is exactly five square feet. Three numbers whose squares are in A.P., with a common difference of $7,$ are $\frac{113}{120}$, $\frac{337}{120}$, $\frac{463}{120}$; and with a common difference of $13$ are $\frac{80929}{19380}$, $\frac{106921}{19380}$, and $\frac{127729}{19380}$. In the case of whole square numbers the common difference will always be divisible by $24,$ so it is obvious that our squares must be fractional. Readers should now try to solve the case where the common difference is $23.$ It is rather a hard nut.

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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