MATH 216T Exam 1 : Part I (In-Class) Solutions 1. (20 pts) A piggy bank contains 24 coins, all of which are nickels (5¢), dimes (10¢) or quarters (25¢). The piggy bank also contains a coin of each denomination. The total value is two dollars. How many combinations of coins are possible? Do not go over all possibilities : you will not receive credit!
Solution : Letn(resp. dandq) be the number of nickels (resp. dimes and quarters) in the piggy bank. Then we have n+d+q= 24 and .05n+.1d+.25q= 2
Multiplying the second equation by 20, we have to find the number of solutions (n, d, q)∈N30of
n+d+q= 24 (1) n+ 2d+ 5q= 40 (2)
Note thatn,dandqare strictly positive since the piggy bank contains a coin of each denomination. We first find all integral solutions. Replacing equation (2) by (2)-(1), we get
n+d+q= 24 (3) d+ 4q= 16 (4)
Since (d, q) = (0,4) is a particular solution of (4), we get that all integral solutions (d, q) of (4) are given by
d= 0 + 4t
q= 4−t wheret∈Z
Substituting this into (3) and solving for n, we find that all integral solutions (n, d, q) are given by n= 20−3t d= 4t q= 4−t wheret∈Z
Since we are interested in solutions withn, d, q >0, we must have
20−3t >0 and 4t >0 and 4−t >0 So
t < 20
3 ≈6.7 and t >0 and t <4
Sincetis an integer, we get that 1≤t≤3. So there are three values fort. This leads to three solutions (n, d, q)∈N3 0. Three
2. (a) (15 pts) Find all the rational points on the hyperbola x2−y2= 1. (b) (5 pts) Find all the integral points on the hyperbolax2−y2= 1.
Solution : (a) We ‘see’ that (1,0) is an integral point on the hyperbola. Hence all the rational points on the hyperbola can be found be intersecting the hyperbola with either a vertical line or a line with rational slope through the point (1,0).
The vertical line through (1,0) has equationx= 1. So we have to solve
x= 1 x2−y2= 1
We easily get that (1,0) is the only solution.
Letq∈Q. The line through (1,0) with slopeq has equationy=q(x−1). So we have to solve
y=q(x−1) (1) x2−y2= 1 (2)
Substituting (1) into (2), we get
x2−(q(x−1))2= 1 or
(1−q2)x2+ 2q2x−q2−1 = 0
If 1−q2= 0 then we get that 2x−2 = 0. Sox= 1 andy= 0. If 1−q26= 0, then we have a quadratic equation inx and we know that x= 1 is a solution. Since the product of the roots ofax2+bx+cis c
a, we get that x= −q2−1 1−q2 1 = q2+ 1 q2−1 Substituting this into (1), we find that
y=q(x−1) =q q2+ 1 q2−1 −1 = 2q q2−1
All the rational points onx2−y2= 1 are given by{(1,0)} ∪
q2+ 1 q2−1, 2q q2−1 :q∈Qandq6=±1
(b) Suppose thatx, y∈Zwithx2−y2= 1. Then
(x−y)(x+y) = 1 Sincex−y, x+y∈Zwe get that
x−y=x+y=−1 or x−y=x+y= 1 So (x, y) = (−1,0) or (x, y) = (1,0).
The only integral points onx2−y2= 1 are (−1,0) and (1,0).
3. (a) (5 pts) Write 101 and 109 as a sum of two squares.
(b) (15 pts) Use (a) to write 44036(= 22·101·109) as a sum of two squares.
You will not receive credit for just writing 44036 as a sum of two squares. You must show your method.
Solution : (a) We easily get that
(b) We used the multiplicative property of the norm inZ[i] to find the following (for alla, b, c, d∈Z):
(a2+b2)(c2+d2) =N(a+bi)N(c+di) =N((a+bi)(c+di)) =N((ac−bd) +i(ad+bc)) = (ac−bd)2+ (ad+bc)2
So using (a), we get
101·109 = (102+ 12)(102+ 32) = (10·10−1·3)2+ (10·3 + 1·10)2= 972+ 402
Multiplying by 4 = 22, we get
44036 = 4·101·109 = 4·972+ 4·402= (2·97)2+ (2·40)2= 1942+ 802
44036 = 1942+ 802
Remark: There is another way of rewriting (a2+b2)(c2+d2) as a sum of two squares by consideringa2+b2=N(a+bi) andc2+d2=N(c−di). This would lead to 44036 = 2062+ 402.
MATH 216T Exam 1 : Part II (Take-Home) Solutions
You must work on these exercises by your self!
You can use any book, calculator, computer, articles on the internet. I am the only person you can ask questions about the exam.
4. Let (x, y, z) be a Pythagorean triple.
(a) (5 pts) Prove that one of the components is divisible by 3. Hint : Consider modulo 3 : what are the squares modulo 3?
(b) (5 pts) Prove that one of the components is divisible by 5 (consider modulo 5). (c) (5 pts) Prove that one of the components is divisible by 4.
Proof : (a) Note that the only squares modulo 3 are {0,1}. Suppose that neither x nory is divisible by 3. Then x2≡y2≡1 mod 3. Hence
z2≡x2+y2≡1 + 1≡2 mod 3 a contradiction since 2 is not a square modulo 3.
Hence eitherxory is divisible by 3.
(b) Note that the only squares modulo 5 are{0,1,4}. Suppose that neitherxnoryis divisible by 5. Thenx2, y2∈ {1,4} mod 5. If x2 ≡y2 ≡1 mod 5 then z2 ≡x2+y2 ≡2 mod 5, a contradiction since 2 is not a square modulo 5. If x2 ≡ y2 ≡4 mod 5 then z2 ≡ x2+y2 ≡ 8 ≡ 3 mod 5, a contradiction since 3 is not a square modulo 5. Hence (x2, y2)∈ {(1,4),(4,1)} mod 5. Soz2≡x2+y2≡1 + 4≡0 mod 5. Hence 5|z2. So 5|z.
(c) We know that there exist d, m, n ∈ N0 such that gcd(m, n) = 1, m > n, not both m and n are odd and (x, y, z)∈ {d(m2−n2,2mn, m2+n2), d(2mn, m2−n2, m2+n2)}. WLOG, (x, y, z) =d(2mn, m2−n2, m2+n2). Then
x= 2mnd. Since eithermor nis even, we see thatxis divisible by 4. 2
5. (20 points) Find all natural numbers aand bwitha2b=b3a.
Solution :
One easily checks that a = 1 ⇐⇒ b = 1. So we may assume that a, b > 1. Hence we can consider the prime factorization ofaandb. We only use the primes that show up in either factorization :
a= n Y i=1 pαi i and b= n Y i=1 pβi i
where p1< p2<· · ·< pn are primes,αi, βi∈Nand not both αi andβi are zero fori= 1,2, . . . , n. Substituting this
into a2b =b3a, we find n Y i=1 p2bαi i = n Y i=1 p3aβi i
Using unique prime factorization, we get
2bαi= 3aβi fori= 1,2, . . . , n (∗)
Soa=kbfor some k∈N0. Sincea2b =b3a, we get that (bk)2b=b3kb Hence (bk)2=b3k and so b3k−2=k2
Ifk= 1 thenb= 1, a contradiction. So k≥2. But thenb3k−2> k2: no solutions. Suppose next that 3a <2b. From (*), we get thatαi< βifori= 1,2, . . . , n. Hence
a= n Y i=1 pαi i divides n Y i=1 pβi i =b
Sob=kafor some k∈Nwithk≥2 (sincea < b). Sincea2b=b3a, we get that
a2ka= (ka)3a Hence a2k= (ka)3 and so a2k−3=k3 Ifk= 2 thena1= 23; soa= 8 andb=ka= 2·8 = 16. Ifk= 3 thena3= 33; soa= 3 andb=ka= 3·3 = 9. Ifk= 4 thena5= 43: no solutions. Ifk= 5 thena7= 53: no solutions. Ifk≥thena2k−3> k3: no solutions.
One easily checks that (8,16) and (3,9) are indeed solutions. So we still have to prove the following:
Letm≥2. Then m3k−2> k2fork= 2,3,4, . . .andm2k−3> k3 fork= 6,7,8, . . . We prove this by induction.
Fork= 2, we get thatm3k−2=m4≥24= 16>4 = 22=k2. Assume thatm3k−2> k2 for somek≥2. Then
m3(k+1)−2=m3·m3k−2>23·k2>4k2= (2k)2>(k+ 1)2
Fork= 6 we get thatm2k−3=m9≥29= 512>216 = 63=k3.
Assume thatm2k−3> k3 for somek≥6. Note that fork >0, we have that
k3≥3k2 ⇐⇒ k≥3 , k3≥3k ⇐⇒ k2≥3 ⇐⇒ k≥√3 and k3≥1 ⇐⇒ k≥1
So
m2(k+1)−3=m2·m2k−3>22·k3=k3+k3+k3+k3≥k3+ 3k3+ 3k+ 1 = (k+ 1)3
6. (20 points) Let pbe a prime with p≡3 mod 4. Prove that there exists exactly one Pythagorean triple such that one of the components is p. Note that we consider the triple (x, y, z) to be the same as (y, x, z).
Proof : Let (x, y, z) be a Pythagorean triple such thatp∈ {x, y, z}. We know that there existd, m, n∈N0such that gcd(m, n) = 1,m > n, not bothmandnare odd and (x, y, z)∈ {d(m2−n2,2mn, m2+n2), d(2mn, m2−n2, m2+n2)}. WLOG, (x, y, z) =d(2mn, m2−n2, m2+n2). We consider three cases: x=p,y =pandz=p.
Suppose first thatx=p. Thenp= 2mnd, a contradiction sincepis odd.
Suppose next thaty=p. Thenp=d(m2−n2) =d(m−n)(m+n). Sincepis a prime andm−n < m+n, we must have thatd=m−n= 1 and m+n=p. Solving for mandn, we get that
m= p+ 1
2 and n=
p−1 2 Substituting this into the formulas for x,y andz, we find that
x= 2mn= 2·p+ 1 2 · p−1 2 = p2−1 2 , y=m 2 −n2=p and z=m2+n2= p −1 2 2 + p+ 1 2 2 =p 2+ 1 2 Note that this is actually a primitive Pythagorean triple.
Suppose finally that z=p. Thenp=d(m2+n2). Sincepis a prime andm2+n2>1, we must have thatd= 1 and m2+n2=p. Sopis a sum of two squares, a contradiction since p≡3 mod 4.
So we proved that there is exactly one Pythagorean triple such thatpis one of the components, namely p2−1
2 , p , p2+ 1
2
7. Leta, b, c∈N0with gcd(a, b) = 1 andc≤ab. Prove that there is at most one couple (x, y)∈N20 withax+by=c. Proof : Suppose there are at least two different non-zero natural solutions toax+by=c. Then the distance between theX andY-intercepts of the lineax+by=c must be strictly bigger than the distance between those two solutions (strictly bigger because the intercepts are not points in N2
0). We’ve seen in class (Proposition 1.3) that the distance between the intercepts is c
ab p
a2+b2 while the distance between two consecutive solutions is √a2+b2. Hence we have c ab p a2+b2>pa2+b2 So c ab >1 orc > ab, a contradiction.
Hence there is at most one couple (x, y)∈N20 withax+by=c. 2
8. Atriangular number is a number of the form n(n+ 1)
2 for somen∈N.
(a) Prove that there are infinitely many triangular numbers that are perfect squares.
(b) Find a triangular number that is a perfect square and is bigger than 1,000,000. Show your work! Proof : (a) Lett∈N. We easily get
t is a triangular number that is a perfect square ⇐⇒ t=m2=n(n+ 1)
2 for some m, n∈N Note that
n(n+ 1)
2 =m
So if t is a triangular number that is a perfect square then (√1 + 8t,√t) is a natural solution of X2−8Y2 = 1. Conversely, let (x, y) be a natural solution ofX2−8Y2= 1. Thenxis odd sincex2= 1 + 8y2 is odd. So x−1
2 ∈N and t:=y2= x 2−1 8 = 1 2 x−1 2 x+ 1 2 = x−1 2 ! x−1 2 + 1 ! 2 is a triangular number that is a perfect square.
We have seen in class that the Pell-Fermat equationX2−8Y2= 1 has infinitely many natural solutions. Hence there are infinitely many triangular numbers that are perfect squares.
(b) We easily get that (3,1) is the fundamental solution ofX2−8Y2= 1. By Theorem 1.33(b), all the solutions of X2−8Y2= 1 are given by ( (3 +√8)k+ (3−√8)k 2 , (3 +√8)k−(3−√8)k 2√8 ! : k= 0,1,2, . . . )
Recall from (a) that y2 is a triangular number that is a perfect square ifx2−8y2 = 1. So we are looking fork∈N with
(3 +√8)k−(3−√8)k
2√8 >
√
1000000 = 1000 We easily get thatk≥5 does the job. So
t:=y2= (3 + √ 8)5−(3−√8)5 2√8 !2 =· · ·= 11892= 1413721
