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Congruent Number Problem Problem Set 1 - Solutions
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(i) Characterization of solutions to Pythagorean Theorem. In this question, we find a
parameterization of the solutions to the equation x2 + y 2 = z 2 .
1. Show that if a prime p divides two of x, y or z, then it must divide the third one
of x, y or z.
Solution: Suppose that p divides both x and y. Then p divides both x2 and y 2
and it also divides the sum which is x2 + y 2 = z 2 . Since p divides the square of a
number, it must also divide z. Similarly, if p divides x and z then it must divide
their squares and hence p divides z 2 − x2 = y 2 . As before, since p divides the
square of a number, it must divide the number itself, namely y. The argument is
similar if p divides y and z.
2. By the previous part, from now on we can suppose that x, y and z share no
common divisors other than ±1. Show now that exactly one of x or y is even.
(Hint: Start by assuming that x and y are both even and showing that x, y, z
share a common divisor. Then assume that both x and y are odd, say x = 2k + 1
and y = 2` + 1 for integers k and ` and then show that powers of 2 on either side
disagree).
Solution: Suppose that both x and y are even. Then z 2 = x2 + y 2 is also even
and hence so is z. Thus x, y, z share 2 as a common divisor which violates our
assumption. Next, suppose that both x and y are odd. Since x and y are odd,
x2 + y 2 = z 2 is even and as before, this implies that z is even. Following the hint,
let x = 2k + 1 and y = 2` + 1 for integers k and ` and suppose that z = 2m for
some integer m. Plugging these in to the equation z 2 = x2 + y 2 gives
z 2 = x2 + y 2
(2m)2 = (2k + 1)2 + (2` + 1)2
4m2 = 4k 2 + 4k + 1 + 4`2 + 4` + 1
3. We’ll assume from now on that y is the even number. Isolate for y 2 , factor and
show that the only common factor of z + x and z − x is 2.
Solution: Let d be a common factor of z − x and z + x. Then d must be a
common factor of their sum and difference, namely d is a common factor of 2z and
d is a common factor of 2x. Since z and x share no common divisor, we conclude
that d must divide 2. Since x and z are odd, we know that 2 is the only common
factor of z − x and z + x.
4. Use unique factorization to conclude that z + x = 2m2 and z − x = 2n2 . Solve for
x, y and z in terms of m and n.
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Solution: From the previous question, if y = 2k for some k, we have (2k)2 =
(z − x)(z + x) and so
z−x z+x
·
k2 =
2
2
where the two factors on the right share no common divisor. Thus, each themselves
must be a perfect square! Hence we can write
z+x
= m2
2
z−x
= n2
2
for integers m and n. Thus z + x = 2m2 and z − x = 2n2 . Adding the two
equations and dividing by 2 gives z = m2 + n2 . Subtracting the two equations
and dividing by 2 gives z = m2 − n2 . Using z + x = 2m2 and z − x = 2n2 with
y 2 = (z − x)(z + x) gives y 2 = 4m2 n2 and so y = ±2mn. (We can ignore the
negative sign by making m and n positive or negative as we see fit. This doesn’t
affect x and z since they involve square terms.)
5. Conclude that all possible characterizations are given by x = d(m2 − n2 ), y =
2dmn, z = d(m2 − n2 ) for integers d, m, n. Also conclude that m and n have
opposite parity (that is, they are not both even or both odd) and that they are
coprime (that is, the only common divisor is 1).
Solution: By no longer assuming that x, y and z do not share a common divisor,
we use the first two parts above to show that they all must have d as a common
divisor. Hence x = d(m2 − n2 ), y = 2dmn, z = d(m2 − n2 ) . Since x/d and z/d
are odd, we see that m and n have opposite parity. If m and n share a common
divisor, then we can absorb the common divisor into the d term since if f divides
m and f divides n then f 2 divides x, y and z so we can divide m and n by f and
include this factor in f .
(ii) In this exercise, we will show that 1 is not congruent by hand. To do this, we will use
the method of infinite descent.
1. First, we will show in parts that x4 + y 4 = z 4 has no integer solution provided
xyz 6= 0. We lose nothing by showing that there are no positive solutions so we
shall do this. First, show that in the equation x2 + y 4 = z 4 as in the previous
exercise, we may suppose that x, y and z share no common factor by reducing the
equation by the common factor if necessary.
Solution: If x and y share a common prime factor p, then as above, we can
show that p must be a factor of z. Similarly if p divides x and z and if p divides
y and z. (I’ll leave it to the interested reader to fill in the details!)
2. Suppose that we have a smallest positive solution (x, y, z) (in the sense of z being
minimal). Now using the previous exercise on the equation x2 + (y 2 )2 = (z 2 )2 , we
may suppose that we are in one of two cases:
i. Case 1: x = 2mn, y 2 = m2 − n2 and z 2 = m2 + n2 .
ii. Case 2: x = m2 − n2 , y 2 = 2mn and z 2 = m2 + n2 .
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In each case, show that we can generate a smaller solution to the equation x2 +y 4 =
z 4 . Hints: For case 1, consider y 2 z 2 . For case 2, argue as follows:
i. Since y 2 = 2mn conclude that m = 2u2 and n = v 2 (or n = 2u2 and m = v 2 ;
we focus on the first case).
ii. Since z 2 = m2 + n2 , we can use exercise 1 to find integers r and s such that
z = r2 + s2 , m = 2rs and n = r2 − s2 . Show that r and s are squares and
that you can find a smaller solution to x2 + y 4 = z 4 .
iii. Thus we decrease forever (infinite descent). We cannot descend forever and
so we have reach a contradiction. We conclude that no solution exists.
Solution: We prove the two cases as outlined above.
In case 1, suppose that x = 2mn, y 2 = m2 − n2 and z 2 = m2 + n2 for integers m
and n (with m ≥ n since y 2 is positive). Then
(yz)2 = y 2 z 2 = (m2 − n2 )(m2 + n2 ) = m4 − n4 .
So (yz, n, m) forms a solution to x2 + y 4 = z 4 . Now, m < z and so we have a
smaller solution.
In case 2, suppose that x = m2 − n2 , y 2 = 2mn and z 2 = m2 + n2 . We proceed
using the hints. Since y 2 = 2mn conclude that m = 2u2 and n = v 2 (or n = 2u2
and m = v 2 - the second case is a good exercise!) Since z 2 = m2 + n2 , we can use
the characterization of Pythagorean triples to find integers r and s with opposite
parity such that z = r2 + s2 , m = 2rs and n = r2 − s2 . Show that r and s are
squares and find a triple of integers that form a smaller solution (in the sense of
a smaller z value) to x2 + y 4 = z 4 .
Hence, we can decrease forever. This is called infinite descent which cannot happen
since there is no infinitely decreasing set of positive integers. This contradicts the
fact that we had a solution to x2 + y 4 = z 4 . Hence this equation has no solution.
3. Conclude that 1 is not a congruent number. Hint:
i. Assume that (a/d)2 +(b/d)2 = (c/d)2 is our rational solution and that ab = 2d2
for positive integers a, b, c, d with d non zero. Any common divisor of a and b
divides both c and d. So assume that a and b have no common divisor.
ii. Since ab = 2d2 , we may write a = 2k 2 and b = `2 for some integers k and `.
Show that 4k 4 + b2 = c2 and conclude that
c+b
= r4
2
c−b
= s4 .
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iii. Solving yields b = r4 −s4 and from before b = `2 . The above exercise disproves
the existence of such integers.
Solution: Assume that (a/d)2 + (b/d)2 = (c/d)2 is our rational solution and that
ab = 2d2 for positive integers a, b, c, d with d non zero. Any common divisor of a
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and b divides c as we did above and it must also divide d which we can see since
ab = 2d2 . So assume that a and b have no common divisor. Since ab = 2d2 , we
may write a = 2k 2 and b = `2 (or vice versa with respect to the power of 2) for
some integers k and `. By some algebra, we have that a2 + b2 = c2 and so
4k 4 + b2 = c2
As before, we factor to see that 4k 4 = (c − b)(c + b). As before again, we see that
the only common divisor of a and b is 2 and so unique factorization tells us that
there are integers r and s such that
c+b
= r4
2
c−b
= s4 .
2
Thus, c + b = 2r4 and c − b = 2s4 . Solving for b and c by adding and subtracting
these two equations gives us that c = r4 + s4 and b = r4 − s4 . From before, b = `2
and so `2 + s4 = r4 . However we just showed that this equation has no integer
solutions! This is a contradiction. Thus, 1 cannot be a congruent number. Phew!
(iii) Using the fact that 1 is not a congruent number, show that
√
2 is irrational.
√
√ 2 √ 2
Solution: Assume towards
a
contradiction
that
2
is
rational.
Then
√
√2 + 2 =
√ √
2 + 2 = 4 = (2)2 and 21 ( 2)( 2) = 1 so 1 is the area of the rational 2 − 2 − 2 right
angle triangle. Thus, 1 is a congruent number, a contradiction.
(iv) Assume the Birch-Swinnerton Dyer Conjecture. Which of the following squarefree
numbers are congruent? Which are not? Which of your proofs did you assume the
Birch-Swinnerton Dyer Conjecture? Can you exhibit a triangle in these cases thus
removing the need to assume the Birch-Swinnerton Dyer Conjecture?
9
10
13
14
55
107
Solution: The numbers 13, 14 and 55 are all congruent numbers. The numbers 9, 10
and 107 are not. Let’s use Tunnel’s Theorem to show one example of each.
For the number 9, the number of solutions to
2x2 + y 2 + 8z 2 = 9
is 10. These are given by (x, y, z) = (±2, ±1, 0), (x, y, z) = (0, ±1, ±1) and (x, y, z) =
(0, ±3, 0). The number of solutions to
2x2 + y 2 + 32z 2 = 9
is 6 as given by (x, y, z) = (±2, ±1, 0) and (x, y, z) = (0, ±3, 0). This is not exactly
double the previous number and so 9 is not congruent by Tunnel’s Theorem. These
solutions are easy to find by just noting that each of x, y and z must be less than 3.
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Assume the Birch Swinnterton-Dyer conjecture. Then for 14, the number of solutions
to
8x2 + 2y 2 + 64z 2 = 14
is the same as the number of solutions to
4x2 + y 2 + 32z 2 = 7
(seen by dividing the previous equation by 2) and this number is 0. Similarly, the
number of solutions to
8x2 + 2y 2 + 16z 2 = 14
is the same as the number of solutions to
4x2 + y 2 + 8z 2 = 7
which is also 0. Thus 14 is congruent assuming the Birch Swinnerton-Dyer conjecture.
(In fact, the right angle triangle with side lengths of 63/6, 8/3 and 65/6 shows that 14
is congruent. How did I find these numbers? Stay tuned to parts 2 and 3!!!)
(v)
1. There are many tricks to show that a number is congruent. Here is one way.
Suppose a, b, c, d are positive integers. Suppose there are nonzero integers x, y, z, w
such that ax2 + by 2 = cz 2 and ax2 − by 2 = dw2 . Show that abcd is a congruent
number. (Taken from: “The Queen of Mathematics: An Introduction to Number
Theory” p. 47). Hint: Show that
(c2 z 4 + d2 w4 )2 ± abcd(4xyzw)2 − 4(cdz 2 w2 ± 2abx2 y 2 )2 = 0.
where the ± signs on either side are the same. Then consider a triangle with side
lengths
U −V
Y
U +V
Y
2X
Y
where
X = c2 z 4 + d 2 w 4
Y = 4xyzw
√
U = X 2 + abcdY 2 = 2(cdz 2 w2 + 2abx2 y 2 )
√
V = X 2 − abcdY 2 = 2(cdz 2 w2 − 2abx2 y 2 )
and write the first hint using these values.
Solution: We show the hint first:
(c2 z 4 + d2 w4 )2 ± abcd(4xyzw)2 − 4(cdz 2 w2 ± 2abx2 y 2 )2
= c4 z 8 + 2c2 z 4 d2 w4 + d4 w8 ± 16abcdx2 y 2 z 2 w2
− 4(c2 d2 z 4 w4 ± 4abcdx2 y 2 z 2 w2 + 4a2 b2 x4 y 4 )
= c4 z 8 + 2c2 z 4 d2 w4 + d4 w8 ± 16abcdx2 y 2 z 2 w2
− 4c2 d2 z 4 w4 ∓ 16abcdx2 y 2 z 2 w2 − 16a2 b2 x4 y 4
= c4 z 8 − 2c2 z 4 d2 w4 + d4 w8 − 16a2 b2 x4 y 4
= (c2 z 4 − d2 w4 )2 − 16a2 b2 x4 y 4
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Now, using the fact that ax2 + by 2 = cz 2 and ax2 − by 2 = dw2 and factoring again,
the above becomes
(c2 z 4 + d2 w4 )2 ± abcd(4xyzw)2 − 4(cdz 2 w2 ± 2abx2 y 2 )2
= (c2 z 4 − d2 w4 )2 − 16a2 b2 x4 y 4
= ((ax2 + by 2 )2 − (ax2 − by 2 )2 )2 − 16a2 b2 x4 y 4
= (((ax2 + by 2 ) − (ax2 − by 2 ))((ax2 + by 2 ) + (ax2 − by 2 )))2 − 16a2 b2 x4 y 4
= ((2by 2 )(2ax2 ))2 − 16a2 b2 x4 y 4
= 16a2 b2 x4 y 4 − 16a2 b2 x4 y 4
=0
Magic! With the hint, we now show that abcd is a congruent number. Using the
two hints above, we see that:
X 2 + abcdY 2 = U 2
and
X 2 − abcdY 2 = V 2
We use the triangle as suggested in the hint. First, notice that it is indeed a right
angle triangle since
2 2 2
U −V
U +V
2X
+
−
Y
Y
Y
2
2
U 2 + 2U V + V 2 4X 2
U − 2U V + V
+
− 2
=
Y2
Y2
Y
2(U 2 + V 2 − 2X 2 )
=
Y
2
2((X + abcdY 2 ) + (X 2 − abcdY 2 ) − 2X 2 )
=
Y2
=0
and thus, since the Pythagorean theorem relationship is satisfied, the triangle is
a right angle triangle. The area of this triangle is given by
U −V
U +V
U2 − V 2
·
=
Y
Y
Y2
(X 2 + abcdY 2 − (X 2 − abcdY 2 )
=
Y2
2
2abcdY
=
Y2
= abcd
and hence abcd is a congruent number. Phew!
2. Use (a, b, c, d) = (1, 1, 1, 7) and find solutions (x, y, z, w) to the two given equations
to show that 7 is a congruent number.
Solution: Take x = 4, y = 3 and plug these into the above equations.
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3. Use (a, b, c, d) = (13, 1, 1, 1) and find solutions (x, y, z, w) to the two given equations to show that 13 is a congruent number.
Solution: Take x = 5 and y = 6 and plug these into the above equations.
4. (Challenging) Use (a, b, c, d) = (1, 1, 1, 23) and find solutions (x, y, z, w) to the two
given equations to show that 23 is a congruent number (the numbers x and y are
rather large in this case!).
Solution: Take x = 312 and y = 266 and plug these into the above equations.
Note that a computer was used to find these numbers (Can you do it in an easier
way?)
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