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Proof step 3 from proof lagrange-lemma-smallest (
previous step
)
prime(j), j ≠ 2 ⊦ ∃ i, i0, j0. 1 + i ^2 + i0 ^2 = j0 * j ∧ 0 < j0 ∧ j0 < j
Lemma
lagrange-lemma