How does 1 not congruent imply Fermat n=4? A natural number is said to be congruent if it is the area of a right triangle with rational sides. I've been told that Fermat actually proved his last theorem for $n=4$ by proving that number 1 is not congruent, but I can't seem to find the connection!
It is probably very easy, thanks in advance.
 A: Assume that $n=1$ is congruent. Then there is a right triangle with rational sides $a,b,c$ such that
$$
(1)\quad a^2+b^2=c^2,
$$
$$
(2) \quad 2ab=4.
$$
Adding $(1)$ and $(2)$ we obtain $(a+b)^2=c^2+4$, substracting them gives $(a-b)^2=c^2-4$. Multiplying the new equations gives
$$
(a^2-b^2)^2=c^4-2^4,
$$
which is a rational solution of $z^2=x^4-y^4$ with $xyz\neq 0$. One can also show that we would obtain a positive integer solution of $z^2=x^4- y^4$, which is a contradiction (because it has none).
The converse implication goes similarly.
Remark: The equation $w^2=u^4-v^4$ is related to Fermat's $x^4+y^4=z^4$, but Fermat considered it only as an auxiliary tool.
A: The key thing to realize is that $1$ being or not being a congruent number is not directly related to the Fermat equation $x^4 + y^4 = z^4$.  It is related to $X^4 + Y^2 = Z^4$, where the second term on the left side is only a square, not a fourth power.  The second equation is what Fermat used, and it is the one you'll see in any account of FLT for exponent $4$.
Any solution to $x^4 + y^4 = z^4$ in positive integers leads to a solution of $X^4 + Y^2 = Z^4$ in positive integers ($X = x, Y = y^2, Z = z$), but not conversely.  So if $X^4 + Y^2 = Z^4$ has no solution in positive integers then $x^4 + y^4 = z^4$ also has no solution in positive integers.  And it's $X^4 + Y^2 = Z^4$ that is closely related to $1$ being or not being a congruent number.
If $X^4 + Y^2 = Z^4$ for positive integers $X$, $Y$, $Z$, then $1$ is a congruent number: $a^2 + b^2 = c^2$ with $(1/2)ab = 1$ using the positive rational numbers $a = Y/(XZ)$, $b = 2XZ/Y$, and $c = (X^4 + Z^4)/(XYZ)$.
Conversely, if $1$ is a congruent number, meaning there are positive rational numbers $a$, $b$, and $c$ such that $a^2 + b^2 = c^2$ and $(1/2)ab = 1$, then give $a$, $b$, and $c$ a common denominator, say $D$: $a = A/D$, $b = B/D$, and $c = C/D$ for positive integers $A, B, C$, and $D$. We get $X^4 + Y^2 = Z^4$ for the rational numbers $X = 2D$, $Y = |A^2 - B^2|$, and $Z = C$. (The integer $Y$ is not $0$, since if it were $0$ then $C^2 = A^2 + B^2 = 2A^2$, making $\sqrt{2} = C/A$ rational, which is false.)
To prove $X^4 + Y^2 = Z^4$ has no solution in positive integers, Fermat used his method of descent to turn a solution $(X,Y,Z)$ in positive integers into another solution $(X',Y',Z')$ of the same equation in positive integers where $Y' < Y$, and repeating that process enough times produces a contradiction since there's no infinite decreasing sequence of positive integers. If you try to run through that descent argument with the equation $X^4 + Y^4 = Z^4$ instead, then the argument doesn't work anymore because the descent process doesn't produce a new solution of the Fermat equation. So it was essential to be working in the broader context of nonsolvability of $X^4 + Y^2 = Z^4$ in order for the descent to work.
You can also relate the nonsolvability of the Fermat equation $x^4 + y^4 = z^4$ with $2$ not being a congruent number by replacing the Fermat equation with the more general equation $X^4 + Y^4 = Z^2$, where the right side is a square rather than a fourth power. The solvability of $X^4 + Y^4 = Z^2$ in positive integers turns out to be equivalent to the solvability of $a^2 + b^2 = c^2$ and $(1/2)ab = 2$ in positive rational $a$, $b$, and $c$, which says $2$ is a congruent number.
Almost every textbook treatment of Fermat's Last Theorem for exponent $4$ proceeds through the nonsolvability in positive integers of $X^4 + Y^4 = Z^2$ rather than $X^4 + Y^2 = Z^4$, which is understandable since the second equation looks a bit awkward compared to the first. So in fact textbook proofs of FLT for exponent $4$ are implicitly based on $2$ not being a congruent number rather than $1$ not being a congruent number. The books almost never explain where the equation $X^4 + Y^4 = Z^2$ comes from.
