Math 480
Test #1 Material
1. If n is any positive integer, then either it is a prime, or else it is divisible by a prime.
Proof by Contradiction: Assume some integer a > 1 does not have a prime divisor. Now the set of
positive integers S = {k ∈ Z and k > 1|k does not have a prime divisor} is nonempty, since a ∈ S.
Thus, by the Well-Ordering property, S has a smallest member, say n. But n is not a prime since
otherwise because n|n, n has a prime divisor, namely n, contradicting the fact that n ∈ S. So n
is composite. Hence ∃x ∈ Z such that 1 < x < n and x|n. Now x ∈
/ S since x < n, the smallest
member of S. Therefore x has a prime divisor,say p. Now p|x and x|n. So p|n and n has a prime
divisor, a contradiction.
2. There is an infinite number of primes.
Suppose there are only finitely many primes, p1 , p2 , . . . , pn . Let N = p1 p2 . . . pn + 1. Since N > 1,
N has a prime divisor (by 13), say q. Then q = pi , for some i such that 1 ≤ i ≤ n. So q|p1 p2 . . . pn .
But q|N . Therefore q|N − p1 p2 . . . pn . So q|1, a contradiction since q > 1.
√
3. If n is not a prime, then it has a prime factor ≤ n.
n > 1 is√composite =⇒ ∃a, b ∈ Z such that 1 < a ≤ b <
√ n.√
√
If a > n then we have the impossibility : n = ab > n n√(since b ≥ a) = n. Hence a ≤ n.
Now a > 1 has a prime divisor, say p (by 13). Then p ≤ a ≤ n and p|n (since p|a and a|n).
√
4. If p is a prime, the p is irrational.
2
√
√
√
Assume p is rational. Then ∃a, b ∈ Z such that p = ab , b 6= 0, gcd(a,b)=1. p = ab =⇒ p = ab2
=⇒ pb2 = a2 =⇒ p|a2 =⇒ p|a (Euclid). Thus ∃ k ∈ Z such that a = pk. Then pb2 = a2
gives pb2 = pk 2 or b2 = pk 2 . Thus p|b2 =⇒ p|b (Euclid). Hence p|a and p|b and gcd(a,b) ≥ p, a
contradiction.
5. If a, b, c is a Pythagorean triple, then one of the integers a or b is even, while the other is odd.
Assume both a and b are even. Then 2|a| and 2|b =⇒ 2|a2 and 2|b2 . So 2|a2 + b2 =⇒ 2|c2 =⇒
2|c|. So, ged(a,b,c)≥ 2. So a,b,c is not a Pythagorean triple.
6. If a, b, c is a Pythagorean triple, then gcd(a,b) = gcd(a,c) = gcd(b,c) = 1.
Assume gcd(a,b)=d > 1. Then ∃ a prime p such that p|d. Then p|a and p|b. =⇒ p|a2 + b2 . So
p|c2 . Thus, p|c. Hendce gcd(a,b,c)≥ p > 1 and a, b, c is not a Pythagorean triple
7. If ab = c2 , where gcd(a,b) = 1, then there exist positive integers a1 , b1 for which a = a21 and b = b21 .
Assume a > 1 and b > 1 and their prime factorizations are
a = pk11 pk22 . . . pkr r , b = q1j1 q2j2 . . . qsjs . Since gcd(a,b)=1, pi ’s and qi ’s are different. Thus, the the
prime factorization of ab is ab = pk11 pk22 . . . pkr r q1j1 q2j2 . . . qsjs .
Let c = ul11 1 ul22 . . . ultt be the prime factorization of c. Since ab = c2 , we have pk11 pk22 . . . pkr r q1j1 q2j2 . . . qsjs =
2lt
1 2l2
u2l
1 u2 . . . ut . Thus, the primes u1 , u2 , . . . , ut are p1 , p2 , . . . , pr , q1 , p2 , . . . , qs in some order and
2l1 , 2l2 , . . . , 2lt are the exponents k1 , k2 , . . . , kr , j1 , j2 , . . . , js . We may assume k1 = 2l1 , k2 =
l
l
2l2 , . . . , kr = 2lr , j1 = 2lr+1 , j2 = 2lr+2 , . . . , j2 = 2ls . Let a1 = pl11 pk22 . . . plrr and a2 = q1r+1 q2r+2 . . . qslt .
Then a = a21 and b = b21 .
8. Let a = 2st, b = s2 − t2 , c = s2 + t2 ; where s, t ∈ Z and s > t > 0
Then
(?).
(a) The formula (?) defines a primitive pythagorean triplet with a even if and only if s and t have
no common factors and are of opposite parity.
(=⇒): Let a,b,c, as given in the formula (?), be a PPT.
Assume gcd(s,t)=d > 1. Then ∃ a prime p such that p|d. Then p|s and p|t. Show that p|2st,
p|s2 − t2 , and p|s2 + t2 .
Also, note that s ≡2 t =⇒ gcd(b,c)≥ 2.
(⇐=) Assume s and t have no common factors and are of opposite parity.
First, verify that a,b, and c, as given in the formula (?), satisfy a2 + b2 = c2 .
Second, show that gcd(a,b,c)=1. Assume gcd(a,b,c)=d > 1. hen ∃ a prime p such that p|d.
Now p|b and p|c. But p|b =⇒ gcd(p,2)=1, since b is odd and p|b and p|c =⇒ p|s2 − t2 and
p|s2 + t2 . Thus p|2s2 and p|2t2 . Since gcd(p,2)=1, p|s2 and p|t2 . Hence, gcd(s,t) > 1.
(b) All primitive pythagorean triplets with a even can obtatined by (?).
Let a,b, and c be a PPT. Since a is even, b and c must be odd. Then c − b and c + b are even.
c+b
a 2
Let u = c−b
2 and v = 2 . Note ( 2 ) = uv and that gcd(u,v)=1 since otherwise d|u + v and
d|u − v and gcd(b,c) > 1
Using the theorem: If ab = c2 , where gcd(a,b) = 1, then there exist positive integers a1 , b1 for
which a = a21 and b = b21 , ∃s, t ∈ Z+ such that u = t2 and v = s2 . Note a = 2st, b = s2 − t2 ,
c = s2 + t2 ; where s, t ∈ Z and s > t > 0. Since gcd(b,c)=1, gcd(s,t)=1. Finally, note s and t
both even or both odd would make both b and c even.
9. Deduce from the description of primitive pythagorean triplets that
a = 2s, b = s2 − 1, c = s2 + 1
with s even give all primitive triplets with c − b = 2.
c-b = 2 =⇒ t=1 and b odd =⇒ s is even.
10. If k is a positive integer show that (2k + 1)2 = 4k 2 + 4k + 1 and hence that the square of an
odd integer is not only odd but leaves the remainder 1 when divided by 4. Using a similar argument, show that the square of an integer not divisible by 3 leaves the remainder 1 when divided by 3.
11. Let m and n be two positive integers. Prove that if m and n have no common factor, then n and
m2 cannot have a common factor. Deduce that if n > 1 and m and n have no common factor, n
cannot divide m2 .
12. Show that if the positive integer k is not the
√ square of another positive integer, it cannot be the
square of a rational number either, so that k is an irrational number.
2
(Hint: if k = m
n2 where m and n are positive integers without a common factor, deduce from
kn2 = m2 that n divides m2 and use the previous exercise to show that n must be 1.)
13. Verify Brahmagupta’s identity:
(x2 − N y 2 ) × (z 2 − N t2 ) = (xz ± N yt)2 − N (xt ± yz)2
= (xz)2 + (N yt)2 ± 2N xzN yt − N ((xt)2 + (yz)2 ± 2xtyz) = x2 z 2 + N 2 y 2 t2 − N (x2 t2 + y 2 z 2 ) =
x2 z 2 +N 2 y 2 t2 −N x2 t2 −N y 2 z 2 = x2 z 2 −N y 2 z 2 +N 2 y 2 t2 −N x2 t2 = z 2 (x2 −N y 2 )−N t2 (x2 −N y 2 ) =
(x2 − N y 2 )(z 2 − N t2 )
14. Verify that if p, q, r, s are all positive integers and (P, Q) = (p, q) ?, (r, s), then P > p, r,
and hence show that (P, Q) is distint from the original two.
Q > q, s
15. Show that
rk > rk−1 , sk > sk−1
and deduce that the (rk , sk ) are all distinct if p, q, r, s are positive integers, using the same method
as in the previous exercise.
16. If x and y is a positive integral solution of X 2 − N Y 2 = m, then
√
N.
x
y
is a rational approximation of
17. Verify that (2, 1) is a solution to X 2 − 3Y 2 = 1. Hence verify that (pk , qk ) are solutions where
(p1 , q1 ) = (2, 1), (pk , qk ) = (2pk−1 + 3qk−1 , pk−1 + 2qk−1 ).
Show that first six solutions are
(2, 1), (7, 4), (26, 15), (97, 56), (362, 209), (1351, 780).
Also starting with the solution (1, 1) to X 2 − 3Y 2 = −2 verify that
(1, 1), (5, 3), (19, 11), (71, 41), (265, 153)
√
are also
solutions to the same equation. Hence deduce the approximation to 3,
√
265
3 < 1351
153 <
780
that were used by Archimedes.
2
(2pk−1 +3qk−1 )2 −3(pk−1 +2qk−1 )2 = 4p2k−1 +9qk−1 )2 +12pk−1 qk−1 −3(p2k−1 +4qk−1
+4pk−1 qk−1 ) =
2
2
pk−1 − 3qk−1 = 1.
Apply the following algorithm to the following three problems.
Find
√
1. p0 , q0 , m0 so that p0 < N < p0 + 1, q0√= 1, m0 = p20 − N
2. Find x1 such that p0 + x1 ≡|m0 | , x1 < N < x1 + |m0 |.
x21 −N
p0 +x1
p0 x1 +N
|m0 | , q1 = |m0 | ,m1 = m0 .
√
If pi , qi , mi = p2i − N qi2 have been found, find xi+1 < N <
x2i+1 −N
+N qi
pi +xi+1 qi
.
Define pi+1 = pi xi+1
,
q
=
,m
=
i+1
i+1
|mi |
|mi |
mi
3. Define p1 =
4.
5.
xi+1 + |mi | pi + xi+1 qi ≡|mi | .
2
2
18. Use the
√ cakravala to find the solution X = 3,2Y = 1 of the equation X − 7Y = 2.
p0 < 7 < p0 + 1 =⇒
√ p0 = 2. q0 = 1, m0 = 2 − 7 = −3.
2 + x1 ≡|−3| 0, x1 < 7 < x1 + | − 3| =⇒ x1 ≡|−3| 1. and x1 = 1 works.
(This much says p0 = 2, q0 = 1 is a solution to X 2 − 7Y 2 = −3, where m0 = −3).
x2 −N
x1 +N
12 −7
0 +x1
, q1 = p|m
,m1 = 1m0 give p1 = 2(1)+7
= 3, q1 = 2+1
Now p1 = p0|m
3
3 = 1,m1 = −3 = 2.
0|
0|
Hence, p1 = 2, q1 = 1 is a solution of X 2 − 7Y 2 = m1 ; that is X 2 − 7Y 2 = 2.
2
2
19. Use the
√ cakravala to find the solution X = 39,2Y = 5 of the equation X − 61Y = −4.
p0 < 61 < p0 + 1 =⇒
p
=
7.
q
=
1,
m
=
7
−
61
=
−12.
0
0
√ 0
7 + x1 ≡|−12| 0, x1 < 61 < x1 + | − 12| =⇒ x1 ≡|−12| 5. and x1 = 5 works.
(This much says p0 = 7, q0 = 1 is a solution to X 2 − 61Y 2 = −12, where m0 = −12).
x2 −N
x1 +N
52 −61
0 +x1
, q1 = p|m
,m1 = 1m0 gives p1 = 7(5)+61
Now p1 = p0|m
= 8, q1 = 7+5
12
12 = 1,m1 = −12 = 3.
0|
0|
Hence, p1 = 8, q1 = 1 is a solution of X 2 −61Y 2 = m1 ; that is X 2 −61Y 2 = 3. Now p1 +x2 q1 ≡|m1 |
√
=⇒ 8 + x2 (1) ≡|3| =⇒ x2 ≡3 1 =⇒ x2 = 1, 4, 7 . . .. But 7 < 61 < 7 + 3. So x2 = 7. Then
pi xi+1 +N qi
, qi+1
|mi |
8+7(1)
+x2 qi
q2 = p1|m
= 3 =
i|
2
X − 61Y 2 = m2 ; that is
pi+1 =
20. Use the
=
x2 −N
pi +xi+1 qi
2 +N q1
,mi+1 = i+1
give So p2 = p1 x|m
|mi |
mi
1|
x22 −N
72 −61
m2 = m1 = 3 = −4 . Hence, p2 = 39, q2 =
2
2
5,
X − 7Y = −4.
=
8(7)+61
3
= 39,
5 is a solution of
cakravala to find the solution X = 41, Y = 5 of the equation X 2 − 67Y 2 = 6.
21. Prove
the two
q identities
q √
√
p
√
2
2
a ± b = (a+ 2a −b) ± (a− 2a −b) .
q √
q √
q √
q √
√
√
(a+ a2 −b)
(a− a2 −b)
(a+ a2 −b)
(a− a2 −b) 2
(a+ a2 −b)
(a− a2 −b)
±
) =
+
±2
=
(
2
2
2
2
2
2
q
q
√
√
√
√
2 −a2 −b)
(a− a2 −b)
(a+ a2 −b) (a− a2 −b)
(a
±2
= a2 + a2 ± 2
=a± b
2
2
2
4
√
(a+ a2 −b)
+
2
√
b
22. The basic approximation in computing square roots is a2 + b ∼ a + 2a
. Prove that
b 2
2
(a ± 2a ) > a ± b.
b2
b
b2
b 2
2
) = a2 + 4a
(a ± 2a
2 ± 2a 2a = a + 4a2 ± b
√
1
1
Use this repeatedly to get 2 ∼ 1 + 31 + 12
− 408
√
3
(Hint: The first application gives 2 > 2. Then 2 = ( 32 )2 − ( 14 ) and so a second application gives
√
3
1
2. A third application gives the required result.)
2 − 12 >
23. Prove the trigonometric identities used in the Archimedes’ approximation.
1
(a) tan(1 θ ) = sin1 θ + tan
θ,
2
q
(b) sin(1 θ ) = 1 + tan21( θ ) .
2
2
24. Writing a(u) = arctan u, and using the formula
tan A+tan B
tan(A + B) = 1−tan
A tan B
tan A−tan B
(and so tan(A − B) = 1+tan
A tan B )
deduce that
u+v
a(u) + a(v) = a( 1−uv
) and
u+v+w−uvw
a(u) + a(v) + a(w) = a( 1−uv−vw−wu
)
25. Use the previous exercise to verify
(i) arctan 12 + arctan 31 = π4
(ii) arctan 12 + arctan 51 + arctan 18 =
1
(iii) 4 arctan 15 − arctan 239
= π4
π
4
1
1
+ arctan 99
= π4
(iv) 4 arctan 51 − arctan 70
37
(Hint: For (iii) and (iv), show that 3 arctan 15 = arctan 55
)
q
p
√
3
3 + 5 + 2 is algebraic by constructing an equation of the 12th degree
7. Show that x =
with integer coefficients satisfied by x.
26. Let a, q
b be positive
with a2 > b. Prove that
q integers
√
√
x = ± a+2 b ± a−2 b
are the roots of the equation
x4 − 2ax2 + b = 0.
√
√
2 −4b
By the quadratic formula, x2 = 2a± 4a
. So x2 = a ± a2 − b. Apply the formula,
2
q √
q √
p
√
2
2
a ± b = (a+ 2a −b) ± (a− 2a −b) .
Replace b by a2 − bq
in the above formula
obtain
q to √
q √
q √
√
√
(a+ a2 −(a2 −b)
(a− a2 −(a2 −b))
(a+ b)
(a− b)
2
2
2
x = a± a − b =
±
.
Thus,
x
=
±
. Hence,
2
2
2
q √
q √ 2
x = ±( (a+2 b) ± (a−2 b) ).
27. Verify the trigonometric identity cos 3θ = 4 cos3 θ − 3 cos θ and hence that x = cos 20◦ satisfies
8x3 − 6x − 1 = 0.