Math 412: Number TheoryLecture 23

Gexin Yugyu@wm.edu

College of William and Mary

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Pythagorean triangle: x2 + y 2 = z2

Defn: a triple of non-zero integers (x , y , z) is called a Pythagoreantriple if x2 + y2 = z2.

Def: a Pythagorean triple (x , y , z) is called a primitive isgcd(x , y , z) = 1.

Lem: if (x , y , z) is a primitive Pythagorean triple, then

I (x , y) = (x , z) = (y , z) = 1,I x is even and y is odd or x is odd and y is even.

Lem: if (r , s) = 1 and rs = t2, then r = m2 and s = n2 for somem.n ∈ N.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Pythagorean triangle: x2 + y 2 = z2

Defn: a triple of non-zero integers (x , y , z) is called a Pythagoreantriple if x2 + y2 = z2.

Def: a Pythagorean triple (x , y , z) is called a primitive isgcd(x , y , z) = 1.

Lem: if (x , y , z) is a primitive Pythagorean triple, then

I (x , y) = (x , z) = (y , z) = 1,I x is even and y is odd or x is odd and y is even.

Lem: if (r , s) = 1 and rs = t2, then r = m2 and s = n2 for somem.n ∈ N.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Pythagorean triangle: x2 + y 2 = z2

Defn: a triple of non-zero integers (x , y , z) is called a Pythagoreantriple if x2 + y2 = z2.

Def: a Pythagorean triple (x , y , z) is called a primitive isgcd(x , y , z) = 1.

Lem: if (x , y , z) is a primitive Pythagorean triple, then

I (x , y) = (x , z) = (y , z) = 1,I x is even and y is odd or x is odd and y is even.

Lem: if (r , s) = 1 and rs = t2, then r = m2 and s = n2 for somem.n ∈ N.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Pythagorean triangle: x2 + y 2 = z2

Defn: a triple of non-zero integers (x , y , z) is called a Pythagoreantriple if x2 + y2 = z2.

Def: a Pythagorean triple (x , y , z) is called a primitive isgcd(x , y , z) = 1.

Lem: if (x , y , z) is a primitive Pythagorean triple, thenI (x , y) = (x , z) = (y , z) = 1,

I x is even and y is odd or x is odd and y is even.

Lem: if (r , s) = 1 and rs = t2, then r = m2 and s = n2 for somem.n ∈ N.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Pythagorean triangle: x2 + y 2 = z2

Defn: a triple of non-zero integers (x , y , z) is called a Pythagoreantriple if x2 + y2 = z2.

Def: a Pythagorean triple (x , y , z) is called a primitive isgcd(x , y , z) = 1.

Lem: if (x , y , z) is a primitive Pythagorean triple, thenI (x , y) = (x , z) = (y , z) = 1,I x is even and y is odd or x is odd and y is even.

Lem: if (r , s) = 1 and rs = t2, then r = m2 and s = n2 for somem.n ∈ N.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Pythagorean triangle: x2 + y 2 = z2

Defn: a triple of non-zero integers (x , y , z) is called a Pythagoreantriple if x2 + y2 = z2.

Def: a Pythagorean triple (x , y , z) is called a primitive isgcd(x , y , z) = 1.

Lem: if (x , y , z) is a primitive Pythagorean triple, thenI (x , y) = (x , z) = (y , z) = 1,I x is even and y is odd or x is odd and y is even.

Lem: if (r , s) = 1 and rs = t2, then r = m2 and s = n2 for somem.n ∈ N.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Thm: (x , y , z) is a primitive Pythagorean triple with y even, if and only if

x = m2 − n2, y = 2mn, z = m2 + n2

where (m, n) = 1 and exactly one of m and n is even.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Fermat’s Last Theorem

Fermat’s Conjecture (1637): For integer n ≥ 3, the equation xn + yn = zn

has no nonzero integer solution.

“ It is impossible to separate a cube into two cubes, or a fourth power intotwo fourth powers, or in general, any power higher than the second, intotwo like powers. I have discovered a truly marvelous proof of this, whichthis margin is too narrow to contain.” (Fermat, 1637)

Fermat did prove the case n = 4. Then in the next 200 years, onlyn = 3, 5, 7 have been proved.

Proved by Andrew Wiles in 1995.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Fermat’s Last Theorem

Fermat’s Conjecture (1637): For integer n ≥ 3, the equation xn + yn = zn

has no nonzero integer solution.

“ It is impossible to separate a cube into two cubes, or a fourth power intotwo fourth powers, or in general, any power higher than the second, intotwo like powers. I have discovered a truly marvelous proof of this, whichthis margin is too narrow to contain.” (Fermat, 1637)

Fermat did prove the case n = 4. Then in the next 200 years, onlyn = 3, 5, 7 have been proved.

Proved by Andrew Wiles in 1995.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Fermat’s Last Theorem

Fermat’s Conjecture (1637): For integer n ≥ 3, the equation xn + yn = zn

has no nonzero integer solution.

“ It is impossible to separate a cube into two cubes, or a fourth power intotwo fourth powers, or in general, any power higher than the second, intotwo like powers. I have discovered a truly marvelous proof of this, whichthis margin is too narrow to contain.” (Fermat, 1637)

Fermat did prove the case n = 4. Then in the next 200 years, onlyn = 3, 5, 7 have been proved.

Proved by Andrew Wiles in 1995.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Fermat’s Last Theorem

Fermat’s Conjecture (1637): For integer n ≥ 3, the equation xn + yn = zn

has no nonzero integer solution.

“ It is impossible to separate a cube into two cubes, or a fourth power intotwo fourth powers, or in general, any power higher than the second, intotwo like powers. I have discovered a truly marvelous proof of this, whichthis margin is too narrow to contain.” (Fermat, 1637)

Fermat did prove the case n = 4. Then in the next 200 years, onlyn = 3, 5, 7 have been proved.

Proved by Andrew Wiles in 1995.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Proposition: There are no nonzero integer solutions of x4 + y4 = z2.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Fermat’s Infinite Descent:

Pierre de Fermat’s method of infinite descent is beautifully illustrated bythe proofs of the above proposition in Number Theory. These areessentially equivalent to Fermat’s own proof, circa 1640, that the area of aPythagorean triangle cannot be a square, which he describe to Huygenssimply by saying ”if the area of such a triangle were a square, then therewould also be a smaller one with the same property, and so on, which isimpossible”.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

rational points on unit circle

Thm: (a, b, c) is a Pythagorean triple if and only if (a/c , b/c) is arational point on unit circle.

To find all rational points on a unit circle, let t be a rational number.Then y = t(x + 1) intersects with x2 + y2 = 1 with a rational point((1− t2/(1 + t2), (2t/(1 + t2)), and every rational point on the unitcircle corresponds to such an intersection.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

rational points on unit circle

Thm: (a, b, c) is a Pythagorean triple if and only if (a/c , b/c) is arational point on unit circle.

To find all rational points on a unit circle, let t be a rational number.Then y = t(x + 1) intersects with x2 + y2 = 1 with a rational point((1− t2/(1 + t2), (2t/(1 + t2)), and every rational point on the unitcircle corresponds to such an intersection.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Sum of (two) squares

Question: For which integer n, does x2 + y2 = n have an integersolution?

Lem: if m and n are solutions, then so is mn.

Lem: every prime p of form 4k + 1 can be written as sum of twosquares.

I There are x , y such that x2 + y2 = mp for some m < p.

I Choose such a pair so that m is minimised. consider a ≡ x , b ≡ y(mod m) with −m/2 < a, b ≤ m/2. Now a2 + b2 ≡ x2 + y2 ≡ 0(mod m), so a2 + b2 = km for some k < m.

I Now (x2 + y2)(a2 + b2) = km2p = (ax + by)2 + (ay − bx)2, and thetwo terms in the last term are multiples of m, sokp = ((ax + by)/m)2 + ((ay − bx)/m)2, a contradiction occurs as onecan show that 0 < k < m.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Sum of (two) squares

Question: For which integer n, does x2 + y2 = n have an integersolution?

Lem: if m and n are solutions, then so is mn.

Lem: every prime p of form 4k + 1 can be written as sum of twosquares.

I There are x , y such that x2 + y2 = mp for some m < p.

I Choose such a pair so that m is minimised. consider a ≡ x , b ≡ y(mod m) with −m/2 < a, b ≤ m/2. Now a2 + b2 ≡ x2 + y2 ≡ 0(mod m), so a2 + b2 = km for some k < m.

I Now (x2 + y2)(a2 + b2) = km2p = (ax + by)2 + (ay − bx)2, and thetwo terms in the last term are multiples of m, sokp = ((ax + by)/m)2 + ((ay − bx)/m)2, a contradiction occurs as onecan show that 0 < k < m.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Sum of (two) squares

Question: For which integer n, does x2 + y2 = n have an integersolution?

Lem: if m and n are solutions, then so is mn.

Lem: every prime p of form 4k + 1 can be written as sum of twosquares.

I There are x , y such that x2 + y2 = mp for some m < p.

I Choose such a pair so that m is minimised. consider a ≡ x , b ≡ y(mod m) with −m/2 < a, b ≤ m/2. Now a2 + b2 ≡ x2 + y2 ≡ 0(mod m), so a2 + b2 = km for some k < m.

I Now (x2 + y2)(a2 + b2) = km2p = (ax + by)2 + (ay − bx)2, and thetwo terms in the last term are multiples of m, sokp = ((ax + by)/m)2 + ((ay − bx)/m)2, a contradiction occurs as onecan show that 0 < k < m.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Sum of (two) squares

Question: For which integer n, does x2 + y2 = n have an integersolution?

Lem: if m and n are solutions, then so is mn.

Lem: every prime p of form 4k + 1 can be written as sum of twosquares.

I There are x , y such that x2 + y2 = mp for some m < p.

I Choose such a pair so that m is minimised. consider a ≡ x , b ≡ y(mod m) with −m/2 < a, b ≤ m/2. Now a2 + b2 ≡ x2 + y2 ≡ 0(mod m), so a2 + b2 = km for some k < m.

I Now (x2 + y2)(a2 + b2) = km2p = (ax + by)2 + (ay − bx)2, and thetwo terms in the last term are multiples of m, sokp = ((ax + by)/m)2 + ((ay − bx)/m)2, a contradiction occurs as onecan show that 0 < k < m.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Sum of (two) squares

Question: For which integer n, does x2 + y2 = n have an integersolution?

Lem: if m and n are solutions, then so is mn.

Lem: every prime p of form 4k + 1 can be written as sum of twosquares.

I There are x , y such that x2 + y2 = mp for some m < p.

I Choose such a pair so that m is minimised. consider a ≡ x , b ≡ y(mod m) with −m/2 < a, b ≤ m/2. Now a2 + b2 ≡ x2 + y2 ≡ 0(mod m), so a2 + b2 = km for some k < m.

I Now (x2 + y2)(a2 + b2) = km2p = (ax + by)2 + (ay − bx)2, and thetwo terms in the last term are multiples of m, sokp = ((ax + by)/m)2 + ((ay − bx)/m)2, a contradiction occurs as onecan show that 0 < k < m.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Sum of (two) squares

Question: For which integer n, does x2 + y2 = n have an integersolution?

Lem: if m and n are solutions, then so is mn.

Lem: every prime p of form 4k + 1 can be written as sum of twosquares.

I There are x , y such that x2 + y2 = mp for some m < p.

I Choose such a pair so that m is minimised. consider a ≡ x , b ≡ y(mod m) with −m/2 < a, b ≤ m/2. Now a2 + b2 ≡ x2 + y2 ≡ 0(mod m), so a2 + b2 = km for some k < m.

I Now (x2 + y2)(a2 + b2) = km2p = (ax + by)2 + (ay − bx)2, and thetwo terms in the last term are multiples of m, sokp = ((ax + by)/m)2 + ((ay − bx)/m)2, a contradiction occurs as onecan show that 0 < k < m.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23

Thm: n is a sum of two squares if and only if each prime factor of n of theform 4k + 3 has an even power.

if n has an odd power of prime factor p of form 4k + 3 andn = x2 + y2.

Let d = (x , y) and consider n/d2 = (x/d)2 + (y/d)2. One can seethat p| n

d2 .

But (x/d , y/d) = 1, so by symmetry we assume that p 6 | xd . Thenthere exists integer z so that (x/d)z ≡ y/d (mod p).

so 0 ≡ (x/d)2 + (y/d)2 ≡ (x/d)2 + (zx/d)2 = (x/d)2(1 + z2)(mod p), thus z2 ≡ −1 (mod p), a contradiction to −1 is anon-quadratic residue modular p.

Gexin Yu gyu@wm.edu Math 412: Number Theory Lecture 23