In The Shadow of Fermat

THE CUBIC EQUATION


TO PROVE π‘Ž3+𝑏3≠𝑐3 {𝑁}.
πΆπ‘œπ‘›π‘ π‘–π‘‘π‘’π‘Ÿ: π‘Ž3+𝑏3=𝑐3 {β„•} {π‘Ž,𝑏,𝑐) π‘Žπ‘Ÿπ‘’ π‘π‘Žπ‘–π‘Ÿπ‘€π‘–π‘ π‘’ π‘π‘œπ‘π‘Ÿπ‘–π‘šπ‘’.

∴ (π‘Ž+𝑏)(π‘Ž2βˆ’π‘Žπ‘+𝑏2)=𝑐3 ∢ (π‘Ž+𝑏) | 𝑐3.

𝐡𝑒𝑑 π‘Ž+𝑏=𝑐+𝑑 ∢ 0<𝑑 ∡ (π‘Ž+𝑏+𝑑)πŸ‘>𝑐3.

∴ (𝒄+𝒅) ∣ π’„πŸ‘.
Let G be the set of factors in {β„•} that divide 𝑐3.
(𝑐+𝑑) | 𝑐3 ⟹ 𝒄 ∣ π’„πŸ‘ ∴ π‘‘βˆˆπΊ ⟹ 𝒅 ∣ π’„πŸ‘. 𝐼𝑓 𝑑 βˆ‰ 𝐺,(𝑐+𝑑) ∀ 𝑑3.

𝐻𝑒𝑛𝑐𝑒 (π‘Ž+𝑏)3=(𝑐+𝑑)3.

π‘Ž3+3π‘Ž2𝑏+3π‘Žπ‘2+𝑏3=𝑐3+3𝑐2𝑑+3𝑐𝑑2+𝑑3.

∴ 3π‘Ž2𝑏+3π‘Žπ‘2=3𝑐2𝑑+3𝑐𝑑2+𝑑3; ∴ 3 ∣ 𝑑3 ⟹ πŸ‘ ∣ 𝒅.

𝐡𝑒𝑑 𝒅 ∣ π’„πŸ‘ ∴ 3 ∣ 𝑐3 ⟹ πŸ‘ ∣ 𝒄.
3π‘Žπ‘(π‘Ž+𝑏)=3𝑐𝑑(𝑐+𝑑)+𝑑3 ⟹ 3π‘Žπ‘(𝑐+𝑑)βˆ’3𝑐𝑑(𝑐+𝑑)=𝑑3

∴ 3(𝑐+𝑑)(π‘Žπ‘βˆ’π‘π‘‘)= 𝑑3;

∴ 32 ∣ πŸ‘(𝑐+𝑑) ⟸ πŸ‘ ∣ 𝒄 & πŸ‘βˆ£π’….

(π‘Žπ‘βˆ’π‘π‘‘) ∣ 𝑑3∢ πŸ‘πŸΒ βˆ£ 𝒄𝒅 ⟹ 32 ∣ (π‘Žπ‘βˆ’π‘π‘‘) ∴ 32 ∣ 𝒂𝒃.

∴ 3 𝑖𝑠 π‘Ž π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿ π‘œπ‘“ {π‘Ž,𝑏,𝑐,𝑑}

π‘‡β„Žπ‘’π‘  33 ∣ (π‘Ž3+𝑏3=𝑐3).
As such, the equation π‘Ž3+𝑏3=𝑐3 can only exist as three terms with an enduring quality that each term is divisible by 3:

equivalent to a proof by infinite descent.
∴ π’‚πŸ‘+π’ƒπŸ‘β‰·π’„πŸ‘ {β„•} 𝑄.𝐸.𝐷.
This methodology for the power of 3 will unfailingly provide the same result, in principle, for any prime power 𝒑β‰₯ 3.

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