2020 IB Extended Essays

2 = 3 − 2 = 8 3 − 2 ≡ − 2 ( 8) ⇔ 2 = − 2 , which is impossible in ℤ . Therefore, must be odd.

And since 3 − 2 = 2 , both and must share the same parity; they must both be odd. Therefore, � + √ 2 , − √ 2 � = 1 ; + √ 2 and − √ 2 are relative primes. 4.3 ± √ 2 as a Cube in ℤ�√− 2 � Since 3 = ∏ 3 ∈ and + √ 2 and − √ 2 are relative primes, there exists a non-empty set ⊂ . . ± √ 2 = ( ∏ ) ∈ 3 . In other words, ± √ 2 is a cube in ℤ�√− 2 � . Therefore, + √ 2 = � + √ 2 � 3 = 3 + 3 2 √ 2 + 3 ( − 2 2 ) − 2 √ 2 3 + √ 2 = � + √ 2 � 3 = ( 3 − 6 2 ) + √ 2 (3 2 − 2 3 ) ∴ � = 3 − 6 2 ⋯ 1 = 3 2 − 2 3 ⋯ From , (3 2 − 2 2 ) = 1 implies that = ±1 and, thus 2 = 1 . Then, since 3 2 − 2 2 = ±1 , too, 3 2 = 2 2 ± 1 = 2 ± 1 = 1 3 . However, it is impossible for integer to satisfy 3 2 = 1 . Therefore, 3 2 = 3 and, thus, 2 = 1 ; = ±1 . From , it can now be deduced that = 3 − 6 2 = ( 2 − 6 2 ) = ±5 . Therefore, Fermat’s Sandwich Theorem can finally be proved. 1. Since = ±5 , 2 = 25 . 2. It then follows that 3 = 2 + 2 = 25 + 2 = 27 , for which the only integer solution is = 3 .

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