Euler's work on Fermat's Last Theorem

n=2 n=4 n=3 E-272

Euler’s Proof of
Fermat’s Last Theorem
(for n = 3)

Lee Stemkoski
Adelphi Univeristy

December 5, 2012

Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 1 / 33

n=2 n=4 n=3 E-272

Outline

1 n=2

2 n=4

3 n=3

4 E-272


n=2 n=4 n=3 E-272

Fun Algebra to Check Thoroughly

If p is prime and p | ab,
then p | a or p | b. (Euclid’s Lemma)
If gcd(r, s) = 1 and r · s = tn ,
then r = un and s = v n .
The set S = {x2 + ny 2 | x, y ∈ Z}
is closed under multiplication.
(a +nb2 )(c2 +nd2 ) = (ac±nbd)2 +n(ad bc)2
2

(Later in this talk, we will consider n = 3.)


n=2 n=4 n=3 E-272

A Babylonian Tablet

Figure: Plimpton 322

Also see:
Euclid, Book X, Lemma 1 - Proposition XXIX

n=2 n=4 n=3 E-272

Proving a Pythagorean Parameterization

Assume a solution exists: x2 + y 2 = z 2 ,
with x, y, z ∈ Z+ , relatively prime.
Some cases to consider:
odd + odd = even
////// +////// = even
even/// even//////////
even + odd = odd
///// +////// =/////
odd/// even//// odd


n=2 n=4 n=3 E-272

(odd)2 + (odd)2 = (even)2

This equation is impossible!

(2m + 1)2 + (2n + 1)2 = (2p)2
4(m2 + m + n2 + n) + 2 = 4p2


n=2 n=4 n=3 E-272

(even)2 + (odd)2 = (odd)2

x2 + y 2 = z 2
x2 = z 2 − y 2
x2 = (z + y)(z − y)
z + y = even = 2p
z − y = even = 2q
x = 2r, y = p − q, z = p + q

p and q: relatively prime, opposite parity


n=2 n=4 n=3 E-272

Finishing it up...

x2 = (z + y)(z − y)
r2 = p · q

gcd(p, q) = 1 implies p = a2 and q = b2

Putting it all together:
√
x = 2r = 2 pq = 2ab

y = p − q = a2 − b 2
z = p + q = a2 + b 2

n=2 n=4 n=3 E-272

The Pythagorean Parameterization

Theorem
If x2 + y 2 = z 2 , with x, y, z relatively prime,
then there exist integers a and b,
relatively prime and with opposite parity,
such that
x = 2ab
y = a2 − b2
z = a2 + b2


n=2 n=4 n=3 E-272

Fermat

Figure: Pierre de Fermat, 1601-1665


n=2 n=4 n=3 E-272

Method of Infinite Descent
Prove: if P (x) is true, then there exists
y < x with P (y) true, where x, y ∈ Z+ .
Obtain an infinite sequence of strictly
decreasing positive integers.
Contradicts the Well Ordering Principle:
S ⊆ Z+ has a smallest element.
Conclude: initial assumption is false.
Useful for showing solutions do not exist.
Fermat’s account of this method: “Relation des
nouvelles dćouvertes en la science des nombres”
e
letter to Pierre de Carcavi, 1659.

n=2 n=4 n=3 E-272

Wanted: Larger Margins

Fermat’s annotation of Bachet’s translation
of Diophantus’ Arithmetica

n=2 n=4 n=3 E-272

A proof

Let X = x2 , Y = y 2 , Z = z 2 , then:

X2 + Y 2 = Z2

X = 2ab, Y = a2 − b2 , Z = a2 + b2

a and b: relatively prime, opposite parity


n=2 n=4 n=3 E-272

The descent

Y = a2 − b2 implies b2 + y 2 = a2
b = 2cd, y = c2 − d2 , a = c2 + d2
c and d: relatively prime, opposite parity

X = 2ab implies x2 = 4cd(c2 + d2 )

cd and (c2 + d2 ): relatively prime

cd = e2 and c2 + d2 = f 2

and c = g 2 , d = h2 ... let the descent begin...

n=2 n=4 n=3 E-272

Recap

Assume there are positive integers such that
X 2 + Y 2 = Z 2 , where X = x2 and Y = y 2
Obtain positive integers such that
c2 + d2 = f 2 , where c = g 2 and d = h2
f is strictly smaller than Z
We obtain an inﬁnite sequence of strictly
decreasing positive integers, which is
impossible; the original assumption was false.


n=2 n=4 n=3 E-272

Euler

Figure: Leonhard Euler, 1707-1783


n=2 n=4 n=3 E-272

Correspondence with Goldbach

196 letters from 1729 to 1764
Goldbach motivates Euler to examine
Fermat’s work
1748 - Euler ﬁrst mentions Fermat’s Last
Theorem
1753 - Euler announces proof for n = 3


n=2 n=4 n=3 E-272

Euler to Goldbach, 13 February 1748


n=2 n=4 n=3 E-272

Euler to Goldbach, 04 August 1753


n=2 n=4 n=3 E-272

Euler’s Algebra (1770)
Exactly one of these three numbers are even.
Case 1: x, y are odd and z is even.
Case 2: y, z are odd and x is even.
Proof of Case 1. Assume that x > y.

x + y = even = 2p and x − y = even = 2q
x = p + q and y = p − q
p and q: positive, relatively prime, opposite parity


n=2 n=4 n=3 E-272

The proof continues

z 3 = x3 + y 3 = (p + q)3 + (p − q)3
= 2p3 + 6pq
= 2p(p2 + 3q 2 )

Note: p2 + 3q 2 is odd.
If g = gcd(2p, p2 + 3q 2 ) > 1 then g is odd;
g | p, so g | 3q 2 ; since g q, we have g = 3.

Therefore, gcd(2p, p2 + 3q 2 ) = 1 or 3.
(Two subcases to consider.)

n=2 n=4 n=3 E-272

gcd(2p, p2 + 3q 2) = 1

2p(p2 + 3q 2 ) = z 3
By earlier fact: 2p = u3 and (p2 + 3q 2 ) = v 3 .
Know: v ∈ S → v 3 ∈ S.
Also: v 3 ∈ S → v ∈ S

p2 + 3q 2 = v 3 = (a2 + 3b2 )3
Since gcd(p, q) = 1, we have:
p = a3 − 9ab2 , q = 3a2 b − 3b3 , gcd(a, b) = 1.

2p = 2a(a + 3b)(a − 3b) = u3

n=2 n=4 n=3 E-272

A descent appears!

2p = 2a(a + 3b)(a − 3b) = u3
2a, (a + 3b), (a − 3b) are relatively prime, so:

2a = α3 , (a + 3b) = β 3 , (a − 3b) = γ 3

α3 = β 3 + γ 3
Move terms if necessary so all terms are positive.
α, β, γ < z, since α3 β 3 γ 3 = 2p < 2p(p2 + 3q 2 ) = z 3
...a smaller positive solution to FLT, n = 3.


n=2 n=4 n=3 E-272

Excerpt #1, Euler’s Algebra


n=2 n=4 n=3 E-272

Excerpt #2, Euler’s Algebra


n=2 n=4 n=3 E-272

Commentarii...

Euler proved FLT(3) in 1770.
Euler proved FLT(3) in 1770, but key steps
were unjustified.
Euler proved FLT(3) in 1770, but key steps
were justified in 1759/1763 (E-272).
The results of E-272 are insufficient to prove
FLT(3).
Euler had a proof of FLT(3) by 1753, but
waited to publish a more polished version.


n=2 n=4 n=3 E-272

Red Flags Revisited

Need to fully justify:
1 If p2 + 3q 2 = v 3
(also gcd(p, q) = 1, and v 3 is odd)
then v = a2 + 3b2 .
2 In this situation, p2 + 3q 2 = (a2 + 3b2 )3
⇒ p = a3 − 9ab2
⇒ q = 3a2 b − 3b2


n=2 n=4 n=3 E-272

The missing link?
E272: Supplementum quorundam theorematum
arithmeticorum... (1759/1763)
Proves properties of numbers of the form x2 + 3y 2 .
Quick tour:
1 If gcd(a, b) = m, then m2 |(a2 + 3b2 )
2 2
2 If 3|(a2 + 3b2 ), then a +3b = n2 + 3m2 .
3
2 2
3 If 4|(a2 + 3b2 ), then a +3b = n2 + 3m2 .
4
4 If P = p + 3q is prime and P |(a2 + 3b2 ),
2 2
2 2
then a +3b = n2 + 3m2 .
P
Corollary: a = 3mq ± np and b = mp ± nq


n=2 n=4 n=3 E-272

5 If Pi = (pi )2 + 3(qi )2 is prime and
2 2
Pi |(a2 + 3b2 ), then a 1+3bk = n2 + 3m2 .
P ...P
6 If A = p2 + 3q 2 is prime and A|(a2 + 3b2 ),
then there exists a similar B < A.
7 All odd prime factors of a2 + 3b2 , when
gcd(a, b) = 1, have the form p2 + 3q 2 .
8 Primes of the form p2 + 3q 2 (except 3)
have the form 6n + 1.
9 Primes of the form 6n + 1
have the form p2 + 3q 2 .


n=2 n=4 n=3 E-272

Red Flag # 1

“If p2 + 3q 2 = v 3
(also gcd(p, q) = 1, and v 3 is odd)
then v = a2 + 3b2 .”

Fully justiﬁed by Euler’s Proposition 7.


n=2 n=4 n=3 E-272

Red Flag # 2

“In this situation, p2 + 3q 2 = (v)3 = (a2 + 3b2 )3
⇒ p = a3 − 9ab2 and q = 3a2 b − 3b2 ”

Can be addressed by Cor. to Euler’s Prop. 4:
Applied to a prime a2 + 3b2 , yields
uniqueness of representation.

(p2 + 3q 2 )(12 + 3 · 02 ) = (a2 + 3b2 )

a = 3 · 0 · q ± 1 · p and b = 0 · p ± 1 · q


n=2 n=4 n=3 E-272

“In this situation, p2 + 3q 2 = (v)3 = (a2 + 3b2 )3
⇒ p = a3 − 9ab2 and q = 3a2 b − 3b2 ”
Repeat for each prime power factor of v;
only one of the two methods of composition
will preserve gcd(a, b) = 1.
Combine prime power factors; many
representations of v.
Must use the same representation to
calculating v 3 to preserve gcd(p, q) = 1.


n=2 n=4 n=3 E-272

Did Euler consider this?

Similar work on sums of two squares
Unusual for Euler to not publish refutation
of FLT for n = 3?
E-255, x3 + y 3 = z 3 + v 3 (pres. 1754)
(one year after letter to Goldbach)
E-256, x2 + cy 2 conjectures (pres. 1753/4)
1755 letter to Goldbach: convinced Fermat
was correct, searching for FLT n = 5 proof
E-272, x2 + 3y 2 (pres. 1759)


Euler's work on Fermat's Last Theorem

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