Is there a Fermat in you?? A solution possible

Reference: Popular Problems and Puzzles in Mathematics, Asok Kumar Mallik, Foundation Books, IISc Press.

Amazon India link:

https://www.amazon.in/Popular-Problems-Puzzles-Mathematics-Mallik/dp/938299386X/ref=sr_1_1?s=books&ie=UTF8&qid=1519414577&sr=1-1&keywords=Popular+problems+and+puzzles+in+mathematics

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:

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

Solution:

We show that the equation y^{3}=x^{2}+2 has only one solution for positive integers x and y, viz., x=5 and y=3. First, we write y^{3}=x^{2}+2=(x+i\sqrt{2})(x-i\sqrt{2}). It can be shown that unique prime factorization occurs in the complex number system a+ ib\sqrt{2}. It can be also argued that since the product of the two complex numbers of that form is a cube, each factor must be a cube. So, we write:

x+i\sqrt{2}=(a+ib\sqrt{2})^{3}=a^{3}-6ab^{2}+i(3a^{2}b-2b^{3})\sqrt{2}.

Equating imaginary parts of both sides, we get

3a^{2}b-2b^{3}=b(3a^{2}-2b^{2})=1.

Thus, b=\pm 1 and 3a^{2}-2b^{2}=\pm 1, both having the same sign. But, with b=-1, one gets a^{2}=1/3, which is not possible. So, we take b=1 when a=\pm 1. Finally, with a=1, x=-5 and with a=-1, x=5. The sign of x is irrelevant as only x^{2} is involved in the original equation. With the solution x=5, we get y=3 as the only set of solution.

Cheers,

Nalin Pithwa.

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