fermats-last-theorem-n4

Fermat’s Last Theorem at $n=4$

Summary: The case $n=4$ of Fermat’s Last Theorem: $x^4+y^4=z^4$ has no nontrivial integer solutions. Stronger statement: $x^4+y^4=z^2$ already has none (so $z^4$ — being a special square — is even more restricted). This is the only exponent Fermat himself proved (in marginal notes on Diophantus); Euler reproduces and amplifies the proof in ch2.0.13-impossibility-biquadrate-sums using infinite descent.

Sources: chapter-2.0.13

Last updated: 2026-05-09

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Statement

For positive integers $x,y,z$:

$x^4+y^4\ne z^4.$

The stronger statement, which implies the above, is:

$\boxed{x^4+y^4\ne z^2.}$

A companion theorem is that $x^4-y^4\ne z^2$ for nontrivial coprime $x,y$.

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Proof Sketch (Infinite Descent)

See ch2.0.13-impossibility-biquadrate-sums for full detail. The structure:

1. Assume a coprime solution $(x,y,z)$ with $x^4+y^4=z^2$.
2. Use the pythagorean-triples parametrization on $(x^2{)}^2+(y^2{)}^2=z^2$: $x^2=p^2-q^2$, $y^2=2pq$.
3. Apply parametrization a second time to $x^2=p^2-q^2\Rightarrow p^2=x^2+q^2$: $p=r^2+s^2$, $q=2rs$.
4. Coprimality forces $r,s,r^2+s^2$ each to be squares: $r=t^2$, $s=u^2$, $r^2+s^2=v^2$.
5. Hence $t^4+u^4=v^2$ — a smaller solution.
6. Descent: contradiction with well-ordering.

This is the textbook example of infinite-descent.

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Why $n=4$ Is “Easy”

Among the FLT exponents, $n=4$ is uniquely tractable for elementary methods because:

• The Pythagorean parametrization can be iterated (legs of one Pythagorean triple are themselves squares, parametrize again).
• Coprimality of three factors ($r,s,r^2+s^2$) forces each to be a square.
• The descent reduces a fourth-power problem to a fourth-power problem of the same shape.

For $n=3$, one needs arithmetic in $\mathbb{Z}[\sqrt{-3}]$ (Euler’s Vollständige Anleitung §243; see fermats-last-theorem-n3 and ch2.0.15-questions-cubes). For higher primes $n$, modern arithmetic geometry is required.

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Implication for FLT at $n=4k$

Once $x^4+y^4=z^2$ is impossible, FLT at any exponent $n=4k$ is automatic: a solution $a^{4k}+b^{4k}=c^{4k}$ would give $(a^k{)}^4+(b^k{)}^4=(c^k{)}^4$, which is also a square — contradicting the case $n=4$.

So to prove FLT in general, it suffices to prove it for odd prime exponents. This is one of Fermat’s structural observations that organized later work (Sophie Germain, Kummer, …, Wiles).

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Historical Notes

• Fermat (c. 1638) sketched the proof of $x^4+y^4\ne z^4$ (and the stronger $\ne z^2$ version) in a marginal note on Diophantus’ Arithmetica. The descent argument is fully outlined in his observations.
• Frenicle de Bessy (1676) published a complete proof.
• Euler (1770) included the proof in Vollständige Anleitung zur Algebra, as reproduced in ch2.0.13-impossibility-biquadrate-sums.
• Wiles (1995) proved FLT for all $n\ge 3$ via the modularity theorem — but the $n=4$ case has remained accessible to undergraduate methods throughout.

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Related pages

• ch2.0.13-impossibility-biquadrate-sums
• ch2.0.15-questions-cubes
• infinite-descent
• fermats-last-theorem-n3
• sum-of-two-cubes
• pythagorean-triples
• indeterminate-analysis