$1$ is not congruent because of Fermat's Last Theorem?

Question

I would like someone to explain something I did not understand. I was reading a page called "nuking the mosquito" where they give very complex proofs for very simple results.

The proof I want to talk about deals with the irrationality of $\sqrt{2}$.

If $\sqrt{2}$ is rational, then there is a right angeled equilateral triangle with sides $(\sqrt{2},\sqrt{2},2)$. The area of this triangle is $1$, and if we assume $\sqrt{2}$ is rational then $1$ is congruent.

Then the writer says "Hence $1$ is congruent. This contradicts Fermat's Last Theorem with exponent $4$"

I don't doubt that $1$ is not congruent, I just don't see the connection to FLT. Would anyone shed some light to what I'm missing?

Answer 1

If $\sqrt{2}$ is rational, then there exists an isosceles triangle has side lengths $(\sqrt{2},\sqrt{2},2)$. This triangle has area $1$; hence $1$ would be a congruent number. This contradicts Fermat with $n=4$; see Theorem $2.1$ in http://www.math.uconn.edu/~kconrad/ross2007/congnumber.pdf for the relation to Fermat.

Answer 2

Pierre de Fermat wrote in the 17th century that: " I have discovered a truly marvelous demonstration of this proposition that this margin is too narrow to contain."

Hypothesize z^3=x^3+y^3.

2^3+4^3=8+64=72. 10^2-1^2-3^3=72.

So 2^3+4^3=10^2-1^2-3^3=[4(4+1)/2]^2 - [2(2-1)/2]^2 - [ (2+1)^3 ].

2^3+8^3=8+512=520. 36^2-1^2-(3^3+4^3+5^3+6^3+7^3)=36^2-1^2-(27+64+125+216+343)=1296-776=520.

So 2^3+8^3=36^2-1^2-(3^3+4^3+5^3+6^3+7^3)=[8(8+1)/2]^2 - [2(2-1)/2]^2 - [(2+1)^3+(2+2)^3+........+(8-2)^3+(8-1)^3]

Rules of the integer x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [ (x+1)^3+(x+2)^3+........+a^3+..........+(y-2)^3+(y-1)^3 ]

So a^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [ (x+1)^3+(x+2)^3+...+(a-1)^3+(a+1)^3+............+(y-2)^3+(y-1)^3 ] - z^3.

Required to have the triad (z,x,y integer) for every (a). However, there exists at least one of (a) such that z, x, y are not the integers.

So z^n =/ x^n+y^n .

Ishtar.