Step
*
1
2
of Lemma
implies-sum-of-two-squares
1. n : ℕ
2. ∀n:ℕn. ∀x:ℕ. (0 < x
⇒ (∃w,y:ℤ. ((n * x * x) = ((w * w) + (y * y)) ∈ ℤ))
⇒ (∃a,b:ℤ. (n = ((a * a) + (b * b)) ∈ ℤ)))
3. x : ℕ
4. ∀x:ℕx. (0 < x
⇒ (∃w,y:ℤ. ((n * x * x) = ((w * w) + (y * y)) ∈ ℤ))
⇒ (∃a,b:ℤ. (n = ((a * a) + (b * b)) ∈ ℤ)))
5. 0 < x
6. w : ℤ
7. y : ℤ
8. (n * x * x) = ((w * w) + (y * y)) ∈ ℤ
9. ¬(n = 0 ∈ ℤ)
10. ¬(∃d:ℕ. ((d | n) ∧ (2 ≤ d) ∧ ((d * d) | n)))
⊢ ∃a,b:ℤ. (n = ((a * a) + (b * b)) ∈ ℤ)
BY
{ (Assert ∀p:Prime. ((p | n)
⇒ (¬((p * p) | n))) BY
Auto) }
1
1. n : ℕ
2. ∀n:ℕn. ∀x:ℕ. (0 < x
⇒ (∃w,y:ℤ. ((n * x * x) = ((w * w) + (y * y)) ∈ ℤ))
⇒ (∃a,b:ℤ. (n = ((a * a) + (b * b)) ∈ ℤ)))
3. x : ℕ
4. ∀x:ℕx. (0 < x
⇒ (∃w,y:ℤ. ((n * x * x) = ((w * w) + (y * y)) ∈ ℤ))
⇒ (∃a,b:ℤ. (n = ((a * a) + (b * b)) ∈ ℤ)))
5. 0 < x
6. w : ℤ
7. y : ℤ
8. (n * x * x) = ((w * w) + (y * y)) ∈ ℤ
9. ¬(n = 0 ∈ ℤ)
10. ¬(∃d:ℕ. ((d | n) ∧ (2 ≤ d) ∧ ((d * d) | n)))
11. ∀p:Prime. ((p | n)
⇒ (¬((p * p) | n)))
⊢ ∃a,b:ℤ. (n = ((a * a) + (b * b)) ∈ ℤ)
Latex:
Latex:
1. n : \mBbbN{}
2. \mforall{}n:\mBbbN{}n. \mforall{}x:\mBbbN{}.
(0 < x {}\mRightarrow{} (\mexists{}w,y:\mBbbZ{}. ((n * x * x) = ((w * w) + (y * y)))) {}\mRightarrow{} (\mexists{}a,b:\mBbbZ{}. (n = ((a * a) + (b * b)))))
3. x : \mBbbN{}
4. \mforall{}x:\mBbbN{}x
(0 < x {}\mRightarrow{} (\mexists{}w,y:\mBbbZ{}. ((n * x * x) = ((w * w) + (y * y)))) {}\mRightarrow{} (\mexists{}a,b:\mBbbZ{}. (n = ((a * a) + (b * b)))))
5. 0 < x
6. w : \mBbbZ{}
7. y : \mBbbZ{}
8. (n * x * x) = ((w * w) + (y * y))
9. \mneg{}(n = 0)
10. \mneg{}(\mexists{}d:\mBbbN{}. ((d | n) \mwedge{} (2 \mleq{} d) \mwedge{} ((d * d) | n)))
\mvdash{} \mexists{}a,b:\mBbbZ{}. (n = ((a * a) + (b * b)))
By
Latex:
(Assert \mforall{}p:Prime. ((p | n) {}\mRightarrow{} (\mneg{}((p * p) | n))) BY
Auto)
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