ELEMENTARY NUMBER THEORY PROBLEM SHEET 9
Exercise 1. Consider a right angled triangle with side lengths x, y, z (z is the
length of the hypotenuse) and denote the radius of the inscribed circle (or incircle)
by r (see diagram below).
By considering the area of the triangle, show that we have
1
1
xy = r(x + y + z)
2
2
Show that, if x, y and z are integers, r is also an integer.
z
x
r
y
Solution 1. We show the result about the area by breaking the triangle into three
smaller triangles (with edges the dashed lines):
z
x
r
r
r
y
The three triangles have area rx/2, ry/2 and rz/2, so the total area of the
triangle (which is also xy/2) is equal to rx/2 + ry/2 + rz/2 = r/2(x + y + z).
Now we have to show that r is an integer. We use the description of Pythagorean
triples from lectures. So we have positive integers d, s, t such that (possibly after
switching x and y), x = d(s2 − t2 ), y = 2dst and z = d(s2 + t2 ). Substituting this
into the equation
1
1
xy = r(x + y + z)
2
2
we get
d2 st(s2 − t2 ) = rd(s2 + st).
Rearranging, we get
r=
d2 st(s2 − t2 )
dt(s2 − t2 )
=
= dt(s − t)
d(s2 + st)
(s + t)
and r is an integer.
Date: Thursday 30th March, 2017.
1
2
ELEMENTARY NUMBER THEORY PROBLEM SHEET 9
Exercise 2.
(1) Fix a positive integer x0 . Show that are only finitely many
integer solutions (y, z) to the equation x20 + y 2 = z 2 . (Hint: consider the
equation x20 = (z + y)(z − y)).
(2) Find three different Pythagorean triples (not necessarily primitive) of the
form (16, y, z).
(3) Find all primitive Pythagorean triples of the form (40, y, z).
Solution 2.
(1) We have x20 = (z + y)(z − y) so z + y has to be a divisor of x20 .
In particular, there are only finitely many possibilities for z + y and z − y.
These determine y and z, so there are only finitely many possibilities for y
and z.
(2) We are going to solve this using the first part, although we could use the
description of primitive Pythagorean triples from lectures. We have 162 =
28 = (z − y)(z + y). We know that z + y > z − y > 0 and z + y, z − y
are congruent mod 2. So the possibilities for z − y are: 2, 22 , 23 and the
corresponding values of z + y are 27 , 26 , 25 . So we get Pythagorean triples
(16, 63, 65), (16, 30, 34), (16, 12, 20).
(3) We use the description of primitive Pythagorean triples from lectures. So
we have positive integers s > t, coprime and different mod 2, with x = 40 =
2st. So st = 20. The possibilities for (s, t) are then (s, t) = (20, 1), (5, 4)
(the conditions on s, t rule out any other possibilities, e.g. (10, 2) is not
allowed since 10 ≡ 2 (mod 2)). So the primitive triples are (40, 399, 401)
and (40, 9, 41).
Exercise 3. Find all Pythagorean triangles whose areas are equal to their perimeter. (Hint: use the description of Pythagorean triples from lectures)
Solution 3. We let the side lengths be x, y, z so the area is 21 xy. Possibly after
swapping x and y, we have integers d, s, t, where s > t are coprime and different
mod 2, such that x = 2dst, y = d(s2 − t2 ), z = d(s2 + t2 ).
We have x + y + z = 2dst + 2ds2 = 2ds(s + t) and 21 xy = d2 st(s2 − t2 ). So if the
area equals the perimeter we have 2ds(s + t) = d2 st(s2 − t2 ). Dividing by ds(s + t)
gives 2 = dt(s − t).
Now we can describe the possible values of d, s, t. If d = 1 we get t(s − t) = 2,
so t = 1, s = 3 is one solution, but t and s are supposed to be different mod 2. So
the only solution with d = 1 is given by t = 2, s = 3. If d = 2 the only solution is
t = 1, s = 2.
So the two Pythagorean triples we get are (5, 12, 13) and (6, 8, 10) and these are
the only possibilities.
Exercise 4. Suppose x, y, z are non-zero integers with y 2 = x3 +xz 4 , and moreover
suppose that x, z are coprime.
(1) Show that x and x2 + z 4 are coprime, and deduce that x and x2 + z 4 are
both squares of integers.
(2) Using the fact that the equation a4 + b4 = c2 has no non-zero integer
solutions, show that y 2 = x3 + xz 4 has no non-zero integer solutions with
x, z coprime.
(3) (*) Show that y 2 = x3 + xz 4 has no non-zero integer solutions.
Solution 4.
(1) Any common divisor of x and x2 + z 4 would divide z 4 , hence
there are no common prime divisors, and x, x2 + z 4 are coprime. We have
x(x2 + z 4 ) = y 2 so both x and x2 + z 4 must be squares.
(2) Suppose x, y, z is a non-zero integer solution with x, z coprime. The previous part gives x = a2 and x2 + z 4 = c2 . So we get a4 + z 4 = c2 which gives
ELEMENTARY NUMBER THEORY PROBLEM SHEET 9
3
a non-zero integer solution to the equation a4 + b4 = c2 . Since there are
no non-zero integer solutions to this equation, we have no non-zero integer
solutions with x, z coprime to y 2 = x3 + xz 4 .
(3) Given a non-zero integer solution (x, y, z) we need to find a non-zero integer
solution with x, z coprime. Suppose p is a prime common divisor of x, z.
Then p3 |y 2 which implies that p2 |y. This implies that p4 |x3 so we also
have p2 |x. This in turn implies that p6 divides the right hand side of
the equation, so p3 divides y. So we get a new non-zero integer solution
( px2 , py3 , pz ). Repeating this process, we eventually get to a solution with
gcd(x, z) = 1. Now we apply the previous part, which says that there are
no such solutions.
Exercise 5. Consider the unit circle C, with equation x2 + y 2 = 1.
(1) Let m be a rational number, and let L be the line passing through (−1, 0)
with slope m. What is the other intersection point between L and C?
(2) Show that taking intersections between L and C as in the previous part
(and letting the rational slope m vary) gives all rational solutions to the
equation x2 + y 2 = 1.
(3) (*) Use the above two parts to give a new proof of the description from
lectures of all primitive Pythagorean triples.
Solution 5.
(1) The equation of the line is y = m(x + 1). Substituting this
into the equation for C gives x2 + m2 (x + 1)2 = 1 which gives
(1 + m2 )x2 + 2m2 x + m2 − 1 = 0
We know that x = −1 is one of the roots of this quadratic polynomial,
so we facto it as
(x + 1)((1 + m2 )x + (m2 − 1))
2
so the other root is x = 1−m
1+m2 . Solving for y gives y = m(x + 1) =
So the other intersection point is
2m
m2 +1 .
1 − m2
2m
,
).
1 + m2 m2 + 1
(2) If we have a rational solution P to x2 + y 2 = 1 (with x 6= −1), then the
line L joining (−1, 0) and P has rational slope (since the coordinates of
P are rational). P is the intersection point between L and C (other than
x = −1).
(3) Suppose (a, b, c) is a primitive Pythagorean triple. We have a2 + b2 = c2 so
2
a 2
+ cb = 1. Now we apply the previous parts: we have
c
a b
1 − m2
2m
=
,
,
c c
1 + m2 m2 + 1
(
for a rational number m = st with gcd(s, t) = 1. Since a, b, c are positive
we have s > t. We get
2
a b
s − t2
2st
,
=
,
.
c c
s2 + t2 s2 + t2
There are now two cases, depending on whether s ≡ t (mod 2) or not.
If s 6≡ t (mod 2) then s2 − t2 and s2 + t2 are coprime, as are 2st and
s2 + t2 (this takes a little bit of work to see). So we get a = s2 − t2 , b =
2
2
s2 +t2
2st, c = s2 + t2 . If s ≡ t (mod 2) then we get a = s −t
2 , b = st, c =
2 .
s−t
Letting u = s+t
and
v
=
then
u,
v
are
coprime,
u
≡
6
v
(mod
2)
and
we
2
2
get a = 2uv, b = u2 − v 2 and c = u2 + v 2 .
4
ELEMENTARY NUMBER THEORY PROBLEM SHEET 9
In both cases, we have shown that our primitive Pythagorean triple has
the form described in lectures.
Exercise 6. (*) Suppose x, y, z are positive integers with x4 − y 4 = 2z 2 , and
moreover suppose that x, y are coprime.
(1) Show that x and y are both odd and gcd(x2 − y 2 , x2 + y 2 ) = 2.
(2) Show that there exist integers a, d such that x2 −y 2 = d2 and x2 +y 2 = 2a2 .
(Hint: first show that (x2 − y 2 ) and (x2 + y 2 )/2 are coprime)
(3) Using the previous part, show that there exist integers b, c such that x+y =
2b2 and x − y = 2c2 .
(4) Show that the integers a, b, c satisfy a2 = b4 + c4 and deduce that there are
no integer solutions to the equation x4 − y 4 = 2z 2 .
Solution 6.
(1) Since x4 − y 4 is even, we deduce that x and y are congruent
mod 2. Since x and y are also coprime they must both be odd. Now let’s
consider x2 − y 2 and x2 + y 2 . If d is a common divisor of x2 − y 2 and x2 + y 2
then d divides 2x2 and 2y 2 . We have gcd(2x2 , 2y 2 ) = 2 gcd(x2 , y 2 ) = 2 since
x and y are coprime.
2
2
2
2
is odd. So x +y
and
(2) Since x2 + y 2 ≡ 2 (mod 4) we deduce that x +y
2
2
(x2 − y 2 ) are coprime. Their product is equal to z 2 , so they are both
squares, and we get the integers a and d.
x−y
2
(3) Since x, y are coprime, x+y
2 and 2 are coprime. Their product is (d/2) ,
so they are both squares, and we get the integers b and c.
2
2
2
2
and b4 + c4 = (x+y) +(x−y)
= a2 . Since there are
(4) We have a2 = x +y
2
4
2
4
4
no non-zero integer solutions to a = b + c , there are no positive integer
solutions to x4 − y 4 = 2z 2 with x, y coprime. In general, if d = 2r d0 is the
gcd of x, y, with d0 odd, then 24r (d0 )4 |2z 2 which implies that d2 |z (since
24r |2z 2 implies 22r |z). So we get a solution (x/d, y/d, z/d2 ) with x/d, y/d
coprime. Since there are no such solutions, we have shown that there are
no non-zero integer solutions to x4 − y 4 = 2z 2 .
Exercise 7. (**)
(1) Show that the equation x4 − y 4 = z 2 has no positive integer solutions.
(Hint: you can use a modification of the argument applied in lectures to
the equation x4 + y 4 = z 2 ).
(2) Using the first part, show that the area of a right angled triangle with
integer length sides cannot be the square of an integer. Deduce that there
is no right angled triangle with area 1 and rational length sides.
Solution 7.
(1) First we reduce to the case that gcd(x, y, z) = 1, exactly as
in the case x4 + y 4 = z 2 . Now we rearrange the equation as y 4 + z 2 = x4 .
So we have a primitive Pythagorean triple (y02 , z0 , x20 ). There are now two
cases to consider.
Firstly, we suppose y0 is even and z0 is odd. So we have positive coprime
integers s > t with s 6≡ t (mod 2) such that y02 = 2st, z0 = s2 − t2 , x20 =
s2 + t2 .
Since (s, t, x0 ) is another primitive Pythagorean triple, we have coprime
integers u, v such that y02 = 2st = 4uv(u2 − v 2 ). Since u, v, u2 − v 2 are all
coprime to each other, they must all be squares. So we get u = a2 , v = b2
and u2 − v 2 = c2 , so a4 − b4 = c2 . We have x0 = u2 + v 2 = a4 + v 2 so
a < x0 . So we have found another positive integer solution (a, b, c) to the
equation with a < x0 .
In the second case, we suppose y0 is odd and z0 is even. So we have
positive coprime integers s > t with s 6≡ t (mod 2) such that y02 = s2 −
ELEMENTARY NUMBER THEORY PROBLEM SHEET 9
5
t2 , z0 = 2st, x20 = s2 + t2 . Multiplying together the first and third equations
we get (x0 y0 )2 = s4 − t4 . So we have a positive integer solution (s, t, x0 y0 )
to the equation with s < x0 .
In both cases, we have found a positive integer solution with a smaller x
value than x0 . Applying Fermat’s descent argument implies that there are
no positive integer solutions.
(2) Suppose we have a right angled triangle with integer side lengths a, b, c,
with a2 + b2 = c2 , and area a square. The area of the triangle is 12 ab.
Dividing by the gcd of a, b (which changes the area by a square factor)
we can assume that the Pythagorean triple (a, b, c) is primitive. So we have
coprime positive integers s, t as usual and 12 ab = st(s2 −t2 ) is a square. Since
s, t, s2 −t2 are coprime to each other, we get that s, t and s2 −t2 are squares.
This implies that we have a positive integer solution to x4 − y 4 = z 2 . By
the previous part, there are no such solutions, so there are no Pythagorean
triangles with square area.
The case of rational side lengths follows immediately: if a right angled
triangle with rational side lengths had area 1, then multiplying by a common denominator of the side lengths would give a Pythagorean triangle
with area a square. But there are no such triangles.
(*) denotes a trickier exercise for those who are interested — it is not essential
for the course. (**) is a longer and/or trickier exercise.