**Fermat’s problem:**

Pierre de Fermat (1601-65), the great French mathematical genius, is described as the “Prince of Amateurs” as he was not a professional mathematician and never published any work during his lifetime. As a civil servant, he served in the judiciary and spent his leisure with mathematics as his hobby. He corresponded with the best of mathematicians of his time. He teased the contemporary English mathematicians, Wallis and Digby, with the following problem, who had to admit defeat.

**Problem:**

**26 is a number that is sandwiched between a perfect square (25) and a perfect cube (27). Prove that there is no other such number.**

Please do try and let me know your solutions. Even partial solutions are welcome.

*Cheers,*

Nalin Pithwa.

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Let n be an integer such that n-1 is a perfect square and n+1 is a perfect cube, i.e. n-1=x^2 and n+1=y^3. Then 2n = x^2 + y^3 and 2 = x^2 – y^3, hence both x and y should be odd numbers. Hence n is an even number, i.e. divisible by 2. Now it remains to prove that other factor of n can be only 13. Let n =2m, then reducing modulo 4 the first equation we will get 0 = 1+y^3 mod 4, i.e y=3 mod 4. Don’t know how to proceed further.

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Let me try if I can develop some further steps…very nice attempt though…

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There is a non-elementary solution I know using the fact that Z[\sqrt{-2}] is a unique factorisation domain, and the fact that y^3 = x^2-2=(x-\sqrt(-2))(x+\sqrt(-2))

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Please do outline this…also for my own and other readers’ benefit…

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This particular solution is very famous, for example on math.se: https://math.stackexchange.com/q/473180

For more theory about Fermat Diophantine equations you can see my summer 2015 report pp. 26 (example 1.7.3) and pp. 45 (example 1.8.1) available here: https://gaurish4math.files.wordpress.com/2015/12/diophantine-equations-gaurish-rev4.pdf

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thanks a lot…

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