ch2.0.13-impossibility-biquadrate-sums

Ch2.0.13 — Of some Expressions of the Form $a{x}^4+b{y}^4$, which are not reducible to Squares

Summary: Euler’s celebrated proof — by infinite descent — that neither $x^4+y^4$ nor $x^4-y^4$ can be a square (for nontrivial integers). This is Fermat’s Last Theorem at exponent $n=4$, the only case Fermat himself proved. Several derived impossibilities follow: $x^4+4y^4$, $x^4-4y^4$, $4x^4-y^4$, $2x^4\pm 2y^4$, and $x^4+2y^4$ are never squares. The chapter contrasts with $x^4-2y^4$, which does become a square in infinitely many cases (e.g., $3^4-2\cdot 2^4=49$).

Sources: chapter-2.0.13

Last updated: 2026-05-09

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The Two Theorems (§202)

For coprime integers $x,y$, neither

$x^4+y^4=z^2\text{nor}{x}^4-y^4=z^2$

has a nontrivial solution. The only solutions are:

• $x^4+y^4=z^2$: $x=0$ or $y=0$.
• $x^4-y^4=z^2$: $y=0$ or $y=x$.

Implication: $x^4+y^4=z^4$ has no nontrivial solutions, since a fourth-power $z^4$ is in particular a square. This is FLT at $n=4$.

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Coprimality Reduction (§203)

Without loss of generality $\gcd (x,y)=1$: any common divisor $d$ would factor out as $d^4$, leaving $p^4\pm q^4=(z/d^2{)}^2$ with $\gcd (p,q)=1$. So one studies only the coprime case.

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Proof That $x^4+y^4=\square$ Is Impossible (§204–205)

Strategy: Infinite Descent

If a coprime solution $(x,y,z)$ exists, construct a strictly smaller coprime solution $(t,u,z^{\prime })$. Iterating must reach the trivial cases — but those cases ($x=0$ or $y=0$) cannot arise from descent because the descent never produces them. Contradiction.

Step-by-step

1. Parities. Both odd is impossible: odd + odd $=4n+2$, not a square. So $x,y$ have opposite parity.

2. Pythagoras. Treating $x^2,y^2$ as legs of a Pythagorean triple with $x^2+y^2=(\text{something})$ — actually $(x^2{)}^2+(y^2{)}^2=z^2$. Write $x^2=p^2-q^2$, $y^2=2pq$, with $p,q$ coprime, opposite parity (using the pythagorean-triples parametrization).

3. Force $p$ odd, $q$ even. If $p$ even: $p^2-q^2\equiv -1(\mathrm{mod}4)$, not a square. So $p$ odd, $q$ even.

4. Second Pythagoras. $x^2=p^2-q^2$ means $p^2=x^2+q^2$. Apply the parametrization again: $p=r^2+s^2$, $q=2rs$, $x=r^2-s^2$, with $\gcd (r,s)=1$.

5. A new biquadrate equation. From $y^2=2pq=4rs(r^2+s^2)$, the quotient $rs(r^2+s^2)$ must be a square. The three factors $r,s,r^2+s^2$ are pairwise coprime, so each is a square: $r=t^2$, $s=u^2$, $r^2+s^2=v^2$. Hence

$\boxed{t^4+u^4=v^2.}$

6. Strict descent. $t,u$ are smaller than $x,y$ (since $x,y$ depend on $t,u$ through fourth powers via the chain $r,s\to p,q\to x,y$). So a smaller coprime solution exists.

7. Conclusion. Iteration produces an infinite strictly decreasing sequence of positive integer solutions — impossible. Hence no nontrivial solution exists.

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Proof That $x^4-y^4=\square$ Is Impossible (§206–208)

A two-case argument:

Case A: Both $x,y$ odd

Set $x=p+q$, $y=p-q$ with one of $p,q$ even, the other odd. Then $x^4-y^4=4pq(2p^2+2q^2)=8pq(p^2+q^2).$ Dividing by 4, $2pq(p^2+q^2)$ must be a square. Coprime factors $\Rightarrow$ each a square: $2p=4r^2$ (so $p=2r^2$), $q=s^2$ (with $s$ odd), and $4r^4+s^4=\square$. Now this $s^4+4r^4=\square$ reduces (by writing $s^4+4r^4$ as sum of two squares $s^4+(2r^2{)}^2$) to making $m^4-n^4$ a square in much smaller numbers. Descent.

Case B: One of $x,y$ even

Say $x$ odd, $y$ even. Then $x^4-y^4=(x^2+y^2)(x^2-y^2)=\square$. Setting $x^2=p^2+q^2$, $y=2pq$, then $x^4-y^4=(p^2-q^2{)}^2$ is automatically a square — but the requirement that the whole formula equal a square forces $p^2-q^2=(p+q)(p-q)=\square$ via the factor $rs(r^2-s^2)$, again leading to a smaller $t^4-u^4=\square$ problem.

Both cases descend; the trivial bases $y=0$ and $y=x$ are not reachable from a chain that started large. Hence impossibility.

Cross-link

§208 observes that the Case-A reasoning, when reformulated, leads back to Case B of $x^4-y^4=\square$ with opposite parity — making the two cases mutually reinforcing.

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Derived Impossibilities (§209)

By embedding into the two main theorems, the following $a{x}^4+b{y}^4=\square$ are also impossible:

Formula	Reduction
$x^4+4y^4$	sum-of-two-squares $\Rightarrow$ $r^4-s^4=\square$ — impossible
$x^4-4y^4$	sum-of-two-squares $\Rightarrow$ $r^4+s^4=\square$ — impossible
$4x^4-y^4$	$y$ must be even; substitute $y=2z$ and divide by 4: reduces to $x^4-4z^4$
$2x^4+2y^4$	divide by 2: $z^4-x^4{y}^4=\square$ — biquadrate difference, impossible
$2x^4-2y^4$	$\gcd (x,y)=1$ forces both odd; substitute $x=p+q,y=p-q$: $16pq(p^2+q^2)=\square$, so $pq(p^2+q^2)=\square$, leading to $p^4+q^4=\square$ — impossible

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$x^4+2y^4$ is Impossible (§210)

A separate descent argument, parallel to but distinct from $x^4+y^4$:

1. $x$ odd (else parity mismatch with the $2y^4$ term).
2. Set $\sqrt{x^4+2y^4}=x^2+2p{y}^2/q$ (the standard rationalization); collecting: $x^2=q^2-2p^2,y^2=2pq.$
3. $q^2-2p^2=\square$ requires $q=r^2+2s^2$, $p=2rs$, hence $x=r^2-2s^2$, and $y^2=4rs(r^2+2s^2)$, so $rs(r^2+2s^2)=\square$.
4. Coprime factors each a square: $r=t^2$, $s=u^2$, $r^2+2s^2=t^4+2u^4=\square$.
5. Descent in smaller numbers; trivial trial confirms no small solution.

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In Contrast: $x^4-2y^4=\square$ Has Infinitely Many Solutions (§211)

This is the striking positive counterpart. The same descent setup leads to two parametrizations because $p^2+2q^2=\square$ admits both $p=r^2-2s^2$ and $p=2s^2-r^2$.

Generation rule (§211.2)

If $2u^4-t^4=\square$, then with $x=t^4+2u^4$, $y=2tu\sqrt{2u^4-t^4}$ we get $x^4-2y^4=\square$.

Bootstrap chain

Seed	Computation	Result
$(t,u)=(1,1)$: $2-1=1=\square$	$x=3$, $y=2$	$3^4-2\cdot 2^4=81-32=49=7^2$
$(t,u)=(1,13)$: $2\cdot 169-1=\dots$, actually $2\cdot 13^4-1=57121=239^2$	$x=1+2\cdot 13^4=57123$, $y=2\cdot 13\cdot 239=6214$	$57123^4-2\cdot 6214^4=\square$
Cycling first solution back: $(t,u)=(3,2)$: $\sqrt{t^4-2u^4}=7$	$x=81+32=113$, $y=2\cdot 3\cdot 2\cdot 7=84$	$113^4-2\cdot 84^4=63473089=7967^2$

The seed $r^2-10s^2=1$ from ch2.0.7-pell-equation-method (Article 140) generates further solutions: $r=13,s=...$, see chapter 2.0.7’s table.

This phenomenon — that $x^4-2y^4$ admits a Pell-like recurrence generating infinitely many solutions — was a major surprise, and it’s why Euler explicitly contrasts it with the $+2y^4$ case.

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Significance

This chapter is the historical apex of Euler’s number-theoretic chapters. It contains:

1. The first published rigorous proof of FLT for $n=4$ (Fermat’s own proof, reconstructed; also covered by Euler in his earlier Vollständige Anleitung zur Algebra).
2. The blueprint for infinite descent — Fermat’s signature method, here in textbook form.
3. The recognition that sign of $c$ matters: $x^4+c{y}^4$ may be uniformly impossible while $x^4-c{y}^4$ admits Pell-driven infinite families.

These propositions form the historical backbone of additive number theory and connect to:

• sum-of-two-cubes — FLT at $n=3$.
• fermats-last-theorem-n4 — modern statement and history.
• infinite-descent — the proof technique.

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Method Summary

Formula	Verdict	Key reduction
$x^4+y^4$	$\ne \square$	descent to smaller $t^4+u^4$
$x^4-y^4$	$\ne \square$	descent to smaller $m^4-n^4$
$x^4+4y^4$	$\ne \square$	reduces to $x^4-y^4=\square$
$x^4-4y^4$	$\ne \square$	reduces to $x^4+y^4=\square$
$4x^4-y^4$	$\ne \square$	reduces to $x^4-4z^4=\square$
$2x^4+2y^4$	$\ne \square$	reduces to $z^4-x^4{y}^4=\square$
$2x^4-2y^4$	$\ne \square$	reduces to $p^4+q^4=\square$
$x^4+2y^4$	$\ne \square$	descent to smaller $t^4+2u^4$
$x^4-2y^4$	$=\square$ infinitely often	bootstrap from seed $(t,u)=(1,1)$, $(3,2)$, …

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Related pages

• ch2.0.11-quadratic-form-factorization
• ch2.0.12-quadratic-form-as-power
• ch2.0.7-pell-equation-method
• fermats-last-theorem-n4
• infinite-descent
• sum-of-two-cubes
• pythagorean-triples
• pell-equation
• indeterminate-analysis