heartless poker revised 2 19 2013 - james madison...
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HEARTLESS POKER
DOMINIC LANPHIER AND LAURA TAALMAN
A. The probabilities, and hence the rankings, of the standardpoker hands are well-known. We
study what happens to the rankings in a game where a deck is used with a suit missing (heartless poker,
for example), or with an extra suit, or extra face cards. In particular, does it ever happen that two or
more hands will be equally likely? In this paper we examine this and other questions, and show how
probability, some analysis, and even number theory can be applied.
1. S, F, F H?
In standard 5-card poker, some hands are obviously better than others. Even the novice player will
easily recognize that Four of a Kind is exciting, while One Pair is comparatively weak. Some players,
however, may occasionally need to be reminded which of Full House, Flush, and Straight is the most
valuable. Without computing probabilities it is not immediately obvious which of these three hands
is the least likely, and thus ranked the highest. The rankings of these three hands in a regular poker
deck with 13 ranks (2-10 and J, Q, K, A) and four suits (spades♠, heartsr, clubs♣, and diamonds
q) are shown in Figure 1.
Straight Flush Full House
4♠
♠4
♠
♠
♠
♠
5r
r5
r
r
r
r
r
6r
r6
r
r
r
r
r
r
7♣
♣7
♣♣
♣
♣♣
♣
♣
8q
q8
q
q
q
q
q
q
q
q <
3r
r3
r
r
r
5r
r5
r
r
r
r
r
9r
r9
r
r
r
r
r
r
r
r
r
Jr
rJ
r
r
Kr
rK
r
r
<
Q♣
♣Q
♣
♣
Qr
rQ
r
r
Q♠
♠Q
♠
♠
4r
r4
r
r
r
r
4q
q4
q
q
q
q
F 1. A Straightis five cards with ranks in sequence. AFlush is five cards of
the same suit. AFull Houseis three cards of one rank and two cards of another
rank. Card images and layout from [2].
Of course it is not difficult to actually compute the necessary probabilities. To build a Full House
you must choose a rank, and then three cards in that rank; thenyou must choose a second rank,1
2 DOMINIC LANPHIER AND LAURA TAALMAN
and then two cards in that rank. Since there are 13 possible ranks and 4 possible suits (hearts,
clubs, spades, diamonds) in a standard deck of cards, the number of ways to build a Full House is(
131
)(
43
)
·(
121
)(
42
)
= 3744. If we divide this number of possible Full House hands bythe total number(
525
)
= 2, 598, 960 of possible hands, we see that the probability of being dealt a Full House from a
standard deck is64165 ≈ 0.00144.
Similar calculations produce the well-known list of the frequencies and probabilities of poker
hands shown in Table 1. Note that each poker hand is represented exactly once in this table; for
example, the number of Flushes excludes those hands that areStraight Flushes, and the number of
Straight Flushes excludes Royal Flushes. The last column isthe ratio of the frequency of a given
poker hand divided by the frequency of the next higher rankedhand. For example, the 6 in the Full
House row asserts that a Full House is 6 times more likely thana Four of a Kind. Similarly, a player
is 17.33 times more likely to get a Four of a Kind than a Straight Flush.
Poker Hand Calculation Frequency Probability Ratio
Royal Flush(
41
)
4 0.00000154 -
Straight Flush(
101
)(
41
)
−(
41
)
36 0.00001385 9
Four of a Kind(
131
)(
44
)
·(
481
)
624 0.00024010 17.333
Full House(
131
)(
43
)
·(
121
)(
42
)
3,744 0.00144058 6
Flush(
135
)(
41
)
−(
101
)(
41
)
5,108 0.00196540 1.364
Straight(
101
)(
41
)5−
(
101
)(
41
)
10,200 0.00392465 1.997
Three of a Kind(
131
)(
43
)
·(
122
)(
41
)254,912 0.02112845 5.384
Two Pair(
132
)(
42
)2·(
111
)(
41
)
123,552 0.04753902 2.25
One Pair(
131
)(
42
)
·(
123
)(
41
)31,098,240 0.42256902 8.889
High Card all other hands 1,302,540 0.50117739 1.186
T 1. Frequency and probability of hands in standard 5-card poker.
The common confusion about the rankings of Full House, Flush, and Straight is understandable;
being dealt a Flush is just 1.997 times more likely than beingdealt a Straight, and being dealt a Full
House is just 1.364 times more likely than being dealt a Flush. To see this graphically, consider the
logarithmically scaled plot of hand frequencies shown in Figure 2. Full House, Straight, and Flush
are clearly clustered together compared to the other hands.Because of this clustering, we will focus
on these three hands in this paper.
HEARTLESS POKER 3
RF SF 4K FH F S 3K 2P 1P HCHand
10
100
1000
10000
100000
1000000
Frequency
F 2. Logarithmic plot of poker frequencies
2. M L: F G P H
Poker is typically played with a standard deck of 52 cards with 13 ranks and 4 suits. There are many
variations of poker by changing various rules such as how many cards are dealt, how many can be
traded, and whether any cards are wild. We could also vary thedeck itself, say by removing certain
cards at the outset.
For example, the game ofHeartless Pokeruses a standard deck with all the hearts removed. The
fact that there are only three suits and 39 cards in the deck changes the probabilities, and perhaps the
rankings, of the different types of hands. With only three suits we would expect itto be much easier
to get a Flush in Heartless Poker than with the usual deck. Moreover, with only three cards of each
rank, we might expect it to be more difficult than usual to get a Straight. In fact, in a moment we will
see that in Heartless Poker the rankings of Flush and Straight are reversed from their usual rankings;
a Heartless Poker Straight is more valuable than a HeartlessPoker Flush, as shown in Figure 3.
Heartless Flush Heartless Straight Heartless Full House
3q
q3
q
q
q
5q
q5
q
q
q
q
q
9q
q9
q
q
q
q
q
q
q
q
q
Jq
qJ
q
q
Kq
qK
q
q
<
4♠
♠4
♠
♠
♠
♠
5♣
♣5
♣
♣
♣
♣
♣
6♣
♣6
♣♣
♣
♣♣
♣
7♣
♣7
♣♣
♣
♣♣
♣
♣
8q
q8
q
q
q
q
q
q
q
q <
Q♣
♣Q
♣
♣
q
q
Q♠
♠Q
♠
♠
4♣
♣4
♣
♣
♣
♣
4q
q4
q
q
q
q
F 3. In Heartless Poker, the relative rankings of the Flush andStraight hands
are reversed.
4 DOMINIC LANPHIER AND LAURA TAALMAN
We could also modify the game of poker by increasing the size of the deck. For example, the
commercially availableFat Packdeck of cards has the standard 13 possible ranks, but each rank
appears ineightsuits (the usual spades, hearts, clubs, and diamonds along with the new suits tridents
È, roses✿, hatchetsO, and dovese). With so many cards in each rank we might expect it to be much
easier to get a Straight, and that is in fact the case. Interestingly, as we will see in a moment, it is
comparatively easy to get a Full House in Fat Pack Poker, but difficult to get a Flush. In other words,
in Fat Pack Poker the rankings of Full House and Flush are reversed from their usual positions, as
illustrated in Figure 4.
Fat Pack Straight Fat Pack Full House Fat Pack Flush
4È
È4
È
È
È
È
5r
r5
r
r
r
r
r
6e
e
6
e
e
e
e
e
e
7♣
♣7
♣♣
♣
♣♣
♣
♣
8✿
✿8
✿
✿
✿
✿
✿
✿
✿
✿ <
Q♣
♣Q
♣
♣
QO
OQ
O
O
Qe
e
Q
e
e
4r
r4
r
r
r
r
4È
È4
È
È
È
È
<
3È
È3
È
È
È
5È
È5
È
È
È
È
È
9È
È9
È
È
È
È
È
È
È
È
È
JÈ
ÈJ
È
È
KÈ
ÈK
È
ÈF 4. When playing with theFat Packdeck, the relative rankings of Flush and
Full House hands are reversed.
Basic counting arguments give the poker hand frequencies inTable 2 for decks withs suits and
r ranks. The entries for Straights and Straight Flushes hold only for r ≥ 6, since forr = 5 the only
possible Straight isA 2 3 4 5 (regardless of whether Ace is considered high or low) and for r < 5 it
is impossible to construct a Straight hand. Note that the sumof the numbers of hands is(
rs5
)
−(
r1
)(
s5
)
,
due to the possibility of obtaining the non-valid poker hand“5 of a kind” whens≥ 5.
The general frequency formulas in Table 2 allow us to prove a small preliminary result:
Theorem 1. Every possible permutation of Straight, Flush, and Full house rankings occurs for some
deck with s suits and r ranks.
Theorem 1 follows immediately by simply applying the frequency formulas in Table 2 to the six
(r, s) examples in Table 3. The first row is the usual 4-suit, 13-rank poker deck, the second row is the
3-suit, 13-rank Heartless Poker deck, and the third row is the 8-suit, 13-rank Fat Pack deck.
HEARTLESS POKER 5
Poker Hand Number of possible handsLeading term
Royal Flush(
s1
)
s
Straight Flush(
r−31
)(
s1
)
−(
s1
)
rs
Four of a Kind(
r1
)(
s4
)(
rs−s1
)
r2s5
Full House(
r1
)(
s3
)(
r−11
)(
s2
)
r2s5
Flush(
r5
)(
s1
)
−(
r−31
)(
s1
)
r5s
Straight(
r−31
)(
s1
)5−
(
r−31
)(
s1
)
rs5
Three of a Kind(
r1
)(
s3
)(
r−12
)(
s1
)2r3s5
Two Pair(
r2
)(
s2
)2(r−21
)(
s1
)
r3s3
One Pair(
r1
)(
s2
)(
r−13
)(
s1
)3r4s4
High Card[
(
r5
)
− (r − 3)][
(
s1
)5− s
]
r5s5
T 2. Frequencies of the possible poker hands for a deck withs suits andr
ranks per suit. The leading terms in the last column are relevant to the next section.
(r, s) Straight Flush Full House Ranking
(13, 4) 10,200 5,108 3,744 S < F < H
(13, 3) 2,400 3,831 468 F < S < H
(13, 8) 327,600 10,216 244,608 S < H < F
(25, 15) 16,705,920 796,620 28,665,000 H < S < F
(30, 7) 453,600 997,353 639,450 F < H < S
(33, 9) 1,771,200 2,135,754 3,193,344 H < F < S
T 3. Examples of frequencies and rankings of Straight (S), Flush (F), and
Full House (H) for decks withssuits andr cards per suit. Note that hands with the
lowest frequencies are ranked highest.
3. A S, F, F H E T?
We now come to the main question of this paper: are there any generalized poker decks for which
Straight, Flush, and Full House hands have the same frequency, either pairwise or all together? We
will handle the latter part of the question in this section and the pairwise question in the next.
6 DOMINIC LANPHIER AND LAURA TAALMAN
From the leading terms in the last column of Table 2 we see thatthe relative rankings of Full
House and Flush will switch infinitely often asr grows and then assgrows, and so on. However, this
is not the case for the relative rankings of Full House and Straight; although Full House is ranked
higher than Straight in standard poker, we will see that thisoccurs only in a finite number of cases.
To aid in our search for pairs (r, s) of rank and suit sizes for which there are ties between hands, we
investigate three curves and their intersections.
Definition 1. For r ≥ 6 and s≥ 1, define the following three curves:
CS F(r, s) : r(r − 1)(r − 2)(r − 4) = 120s4
CS H(r, s) : r(r − 1)s(s− 1)(s− 2) = 12(r − 3)(s+ 1)(s2 + 1)
CFH (r, s) : r(r − 1)(r − 2)(r − 3)(r − 4) = 120(r − 3)+ 10r(r − 1)s(s− 1)2(s− 2)
The equationCS F(r, s) is obtained by setting the frequencies of Straight (S) and Flush (F) from
Table 2 equal to each other and simplifying. This means that apoint (r, s) is on the curveCS F(r, s)
if and only if the Straight and Flush hands have equal frequency in a generalized poker deck withr
ranks ands suits. Similarly,CS H describes the decks for which Straight and Full House (H) have
equal frequency, andCFH describes the decks for which Flush and Full House have equalfrequency.
Figure 5 shows the graphs ofCS F(r, s), CS H(r, s), andCFH (r, s) for 6 ≤ r ≤ 50, 1≤ s ≤ 20. The
decreasing curve isCS H(r, s), the steeper of the two increasing curves isCS F(r, s), and the remaining
curve isCFH (r, s). The regions bounded by the curves represent where one poker hand outranks
another; for example, the six dots in the figure represent thesix examples in Table 3.
The point in Figure 5 at which the three curves meet represents a value (r, s) for which the fre-
quency formulas forS, F, andH are all equal. Our first result is to show that there is only onesuch
point, and that it has non-integer coordinates.
Theorem 2. If r ≥ 6 and s≥ 1 then there exists exactly one point(r, s) that simultaneously satisfies
CS F(r, s), CS H(r, s), and CFH (r, s), and this solution has22 < r < 23 and6 < s < 7. Consequently,
there is no generalized poker deck with r ranks and s suits forwhich Straight, Flush, and Full House
together have equal ranking.
Proof. It suffices to show that there is only one (r, s) that simultaneously satisfiesCS H(r, s) and
CS F(r, s). Implicitly differentiatingCS F(r, s) gives
480s3 dsdr=
d fdr,
HEARTLESS POKER 7
10 20 30 40 50
5
10
15
20
F 5. CS F(r, s), CS H(r, s), andCFH (r, s) intersect at exactly one point.
wheref (r) = r(r − 1)(r − 2)(r − 4). Since clearlyd fdr > 0 for r ≥ 6, and sinces≥ 1 by hypothesis, we
havedsdr > 0 for points (r, s) onCS F.
For g(r) = r(r−1)12(r−3) theng′(r) = r2−6r+3
12(r−3)2 > 0 for r ≥ 6. Thusg(r) is increasing for suchr and so
g(r) ≤ g(20)= 1.86 for 6≤ r ≤ 20. Now, for (r, s) onCS H(r, s) we haveg(r) = h(s) = (s+1)(s2+1)s(s−1)(s−2) and
soh(s) ≤ 1.86 for such (r, s) with 6 ≤ r ≤ 20. A straightforward calculation shows thath(s) ≤ 1.86
implies 6.58s2 + 1 ≤ 0.86s3 + 2.72s. Clearly this inequality doesnot hold for s = 1 ands = 6 and
these two polynomials are increasing and concave up fors≥ 1. Now, the derivative of 0.86s3+2.72s
is smaller than the derivative of 6.58s2 + 1 for 1 ≤ s ≤ 4.893 and larger for 4.893 ≤ s. Further,
6.58s2 + 1 > 0.86s3 + 2.72s at s = 4.893. Therefore, as 6.58s2 + 1 > 0.86s3 + 2.72s at s = 6 also,
then it must hold for all 1≤ s ≤ 6. It follows thath(s) > 1.86 for s ≤ 6. So we must haves > 6 for
r ≤ 20. However, if a point (r, s) with r ≤ 20 is also on the curveCS F(r, s) then we must have
s=
(
r(r − 1)(r − 2)(r − 4)120
)1/4
≤(
20 · 29 · 18 · 16120
)1/4
= 5.49.
So we must haver ≥ 20. From the same argument in the previous line it follows that we must also
haves≥ 5.49.
8 DOMINIC LANPHIER AND LAURA TAALMAN
Implicitly differentiatingCS H(r, s) gives
dsdr=
12(s+ 1)(s2 + 1)− (2r − 1)s(s− 1)(s− 2)r(r−1)(s−1)(s−2)+ r(r−1)s(s−2)+ r(r−1)s(s−1)− 12(r−3)(s2+1)− 24(r−3)(s+1)s
.
For s ≥ 5 we have 3(s+ 1) < 396 (s− 2) and 4(s2 + 1) < 6s(s+ 1). It follows that the numerator of
ds/dr is less than 0 fors≥ 5. For the denominator, settingj(s) = 3s2+2s+13s2−6s+2 then j′(s) = −24s2+6s+8
(3s2−6s+2)2 < 0
for s ≥ 5. Thus j(s) is decreasing and soj(s) ≤ j(5) = 1.83 for s ≥ 5. Recall thatg(r) from the
previous paragraph is at least 1.86 forr ≥ 20 and so we must haveg(r) > j(s) for r ≥ 20 ands ≥ 5.
Thus r(r−1)12(r−3) >
3s2+2s+13s2−6s+2 and, rewriting, this is
r(r − 1) ((s− 1)(s− 2)+ s(s− 2)+ s(s− 1)) > 12(r − 3)(
(s2 + 1)+ 2s(s+ 1))
.
Thus the denominator ofds/d f is positive for suchr andsand sodsdr < 0 for points (r, s) onCS H(r, s).
ThereforeCS F(r, s) andCS H(r, s) have at most one intersection point, and this clearly occurs at
the point indicated in Figure 5. �
4. N T: A D E P
Our main result is that Straight, Flush, and Full House have distinct frequencies in any generalized
poker game. The proof primarily uses elementary number theory and some elementary analysis.
However, a fairly deep result in Diophantine equations is also necessary.
Theorem 3. There is no pair of integers(r, s) with r ≥ 6 and s≥ 1 that satisfies any of CS F(r, s),
CS H(r, s), and CFH (r, s). Consequently, there is no generalized poker deck with r ranks s suits for
which any pair of Straight, Flush, or Full House have the sameranking.
Proof. We will show that each of the three equations has no solution with integersr ≥ 6 ands ≥ 1,
starting withCFH (r, s). Suppose that (r, s) is an integral solution ofCFH (r, s), and consider the
equationCFH (r, s) modulor − 1. This gives 0≡ −240 (modr − 1), which implies thatr − 1 divides
240= 24 · 3 · 5. On the other hand, considering equationCFH (r, s) modulor we have 0≡ −360 (mod
r), which implies thatr divides 360= 23 · 32 · 5. Sincer andr − 1 are relatively prime this means
thatr(r − 1) divides 24 · 32 · 5 = 720. Thereforer(r − 1) ≤ 720 and so
r2 − r − 720≤ 0.
By the quadratic formula we haver ≤ 27. The only integers 6≤ r ≤ 27 for which r | 360 and
(r − 1) | 240 arer = 6 andr = 9.
HEARTLESS POKER 9
For r = 6, equationCFH (r, s) becomes
760− 360+ 300s(s− 1)2(s− 2),
which gives 6/5 = s(s− 1)2(s− 2), and that equation that has no solution for integrals. For r = 9,
equationCFH (r, s) is
15120= 720+ 720s(s− 1)2(s− 2),
which simplifies to 20= s(s− 1)2(s− 2). Clearly a positive integral solution to this equation must
satisfys > 2. If f (s) = s(s− 1)2(s− 2) then f ′(s) is positive, and thusf (s) is increasing, fors > 2.
Since f (3) = 12 andf (4) = 72, there is no integersso thatf (s) = 20.
We now turn our attention toCS H(r, s). We will show that there are only a finite number of positive
integral pairs (r, s) for which a Full House outranks a Straight. In particular, we will show that
r(r − 1)s(s− 1)(s− 2) > 12(r − 3)(s+ 1)(s2 + 1)
for r ≥ 23 ands≥ 7, and then later investigate the cases for smallerr ands. First note that forr ≥ 23
we haver(r−1) > 24(r−3) because by the quadratic formular2−25r+72> 0 for r > 25+√
3772 ≈ 22.2.
It now suffices to prove that 2s(s− 1)(s− 2) > (s+ 1)(s2 + 1), or equivalently, that
(s+ 1)(s2 + 1)s(s− 1)(s− 2)
= 1+4s2 − s+ 1
s3 − 3s2 + 2s< 2.
We haves3 − 3s2 + 2s < 4s2 − s+ 1 exactly wheng(s) = s3 − 7s2 + 3s− 1 > 0. This follows for
s≥ 7 becauseg′(s) is positive fors≥ 5 andg(7) = 20> 0.
To prove thatCS H(r, s) has no integral solutions it now suffices to show that the equation does not
hold for the casess ∈ {1, . . . , 6} andr ∈ {6, . . . , 22}. The casess = 1, 2 are obvious. Fors = 3 the
equationCS H becomes
r(r − 1)6= 80(r − 3),
or equivalently,r2− 81r + 240= 0, which has no integral solutions. The casess= 4, 5, 6 are similar.
Note that forr = 6 the equationCS H becomes
5s(s− 1)(s− 2) = 6(s+ 1)(s2 + 1),
or equivalently,s3 + 21s2 − 4s+ 6 = 0, which has no integral solutions for anysbecauses3 + 4s2 −
4s+ 6 > 0 for all s. The casesr = 7, . . . , 22 can be done similarly.
Finally, we must prove that the equationCS F(r, s) given byr(r − 1)(r − 2)(r − 4) = 120s4 has
no integral solutions withr ≥ 6 ands ≥ 1, i.e., that Straight and Flush can never be tied in any
10 DOMINIC LANPHIER AND LAURA TAALMAN
generalized (r, s) poker game. This is where we employ a classical result in Diophantine equations.
If ( r, s) is a pair of positive integers withr(r − 1)(r − 2)(r − 4) = 120s4 then we can write
r = αu4, r − 1 = βv4, r − 2 = γz4, and r − 4 = δw4
whereα, β, γ, δ, u, v, z,w are positive integers andα β γ δ = 120 = 23 · 3 · 5. This gives rise to the
following set of Diophantine equations:
(1) αu4 − βv4 = 1, βv4 − γz4 = 1, αu4 − γz4 = 2, γz4 − δw4 = 2, βv4 − δw4 = 3, αu4 − δw4 = 4.
We will show that no integerα dividing 23 · 3 · 5 and satisfying all of the above conditions can exist.
First note that if 23 | α then the third equation in (1) implies thatγz4 is even. Sinceαγ | 23 · 3 · 5
this means thatγ is odd; thereforez4 is even, and thusz is even. Thenz4 = 24z′4 for somez′ ∈ Z.
Therefore the third equation in (1) implies that
2 = αu4 − γz4 = 23α′u4 − 24γz′4
and so 1= 22α′u4 − 23γz′4, which gives a contradiction since the right hand side of theequation is
even. Therefore it follows that 23 ∤ α.
In a similar way, if 22 | α then it follows thatγ must be even and henceβ andδ are odd.
To tackle the rest of the cases, we repeatedly apply a couple of ingredients. The first is the
observation that for anya ∈ Z we have
(2) a4 ≡ 1 or 0 (mod 3) anda4 ≡ 1 or 0 (mod 5).
The second ingredient is a fairly classical result in Diophantine equations. The following theorem
was shown in [3] and generalized in [1].
Theorem 4(Bennett/DeWeger/Ljunggren). Let n,m be integers. The equation
|nx4 −my4| = 1
has at most one solution in positive integers x and y.
Suppose that 3| α, so thenα ∈ {3, 2 · 3, 22 · 3, 3 · 5, 2 · 3 · 5, 22 · 3 · 5}. Taking the first of the
equations of (1) modulo 3 we getαu4 − βv4 ≡ 1 (mod 3) and thus−βv4 ≡ 1 (mod 3). This implies
that−β ≡ 1 (mod 3) and soβ ≡ 2 (mod 3). We can do the same thing to the equationsαu4−δw4 = 4
andαu4 − γz4 = 2 and we getδ ≡ 2 (mod 3) andγ ≡ 1 (mod 3). Note also thatβv4 − δw4 ≡ 3
HEARTLESS POKER 11
(mod 5). Combining this with the fact thatα β γ δ = 23 · 3 · 5, we see that the ordered quadruple of
integers (α, β, γ, δ) can only be one of
(3) (3, 23, 1, 5), (3, 22 · 5, 1, 2), (22 · 3, 5, 1, 2), (3, 5, 22, 2)
From the equations in (1), the first of these quadruples implies thatz4 − 5w4 = 2. Taking this
modulo 5 givesz4 ≡ 2 (mod 5) which gives a contradiction. The second quadruple gives 3u4 −
22 · 5v4 ≡ 1 (mod 5) which implies 3u4 ≡ 1 (mod 5) and sou4 ≡ 2 (mod 5) which again gives
a contradiction. We previously showed that if 22 | α thenγ must be even, so we can immediately
eliminate the third quadruple.
Theorem 4 implies that the equationβv4 − γz4 = 1 has at most one solution forv, z positive
integers. For the fourth quadruple,β = 5 andγ = 22 and we have the solutionv = z = 1. So by
Theorem 4 there can be no more solutions withv, z positive integers. Now,v = z= 1, β = 5, γ = 22
imply r = 6 and then equationCS F(r, s) simplifies tos4 = 2 which has no integral solutions. So we
cannot have this case and this eliminates the fourth quadruple in (3). Thus we have that 3∤ α.
Now suppose that 5| α. Thenαu4 − βv4 ≡ 1 (mod 5) means that−βv4 ≡ 1 (mod 5) and from
(2) we have thatβ ≡ 4 (mod 5). Therefore the only possibilities forβ are 4 and 24. Note that
αu4 − βv4 = 1 implies (α, β) = 1 and as 3∤ α we must haveα = 5. If β = 4 then|αu4 − βv4| = 1 has
the solutionu = v = 1 and by Theorem 4 there are no other solutions withu, v positive integers. Thus
r = αu4 = 5 and equationCS F(r, s) simplifies to 2s4 = 1 which has no integral solutions. Therefore
β , 4, and thusβ = 24 is the only possible value forβ. This means that the only possible values of
the constants are given by the quadruple (α, β, γ, δ) = (5, 23 · 3, 1, 1). Then the fourth equation in (1)
becomesz4−w4 = 2. But by (2), we know thatz4,w4 can only be 0 or 1 modulo 5 and soz4−w4 can
only be 0, 1, or 4 modulo 5. This contradicts (2). Hence we musthave that 5∤ α.
We are now left with the cases whereα is one of 1, 2, or 22. If α = 22 then as mentioned
previously,γ is even andβ, δ are odd. If|α − β| = 1 then by Theorem 4 we must haveu = v = 1
and sor = αu4 = 4. As r ≥ 5, this cannot occur and soβ , 3, 5. Thus the only possibilities for
β are 1 and 15. But ifβ = 15 thenαu4 − βv4 ≡ 1 (mod 5) gives 4u4 ≡ 1 (mod 5) which again
contradicts (2). So we must haveβ = 1; but now we have
αu4 − βv4 = 1 = 4u4 − v4 = (2u2 − v2)(2u2 + v2),
or equivalently, 2u2 + v2 = 1, which has only the solutionu = 0, v2 = 1. This implies thatr = 0 and
thus gives a contradiction. Thereforeα , 22.
12 DOMINIC LANPHIER AND LAURA TAALMAN
Now for the caseα = 2. If 3 | β then consider the equation 2u4 − βv4 = 1 modulo 3. We get
2u4 ≡ 1 (mod 3) and sou4 ≡ 2 (mod 3) which gives a contradiction. Thus 3∤ β and in a similar
way, by taking the equation modulo 5, we have that 5∤ β. We can do the same thing forδ using
the equation 2u4 − δw4 = 4 and get 3, 5 ∤ δ. Thus we must have that 3· 5 | γ. From the equation
βv4 − γz4 = 1 we get thatβv4 ≡ 1 (mod 3) andβv4 ≡ 1 (mod 5). Thusβ ≡ 1 (mod 3) andβ ≡ 1
(mod 5). But as 3, 5 ∤ β, thenβ can only be 1 or 2 or 22 and onlyβ = 1 is congruent to 1 modulo
5. But then the equationαu4 − βv4 = 1 becomes 2u4 − v4 = 1 which by Theorem 4 only has one
solution, namelyu = v = 1. This impliesr = 2, but by hypothesis we haver ≥ 5. So we must have
α , 2.
Finally, if α = 1 then consider the equationu4 − γz4 = 2. Sinceu4 ≡ 0 or 1 modulo 3 or 5, this
means thatγz4 ≡ 1 or 2 modulo 3 andγz4 ≡ 3 or 4 modulo 5. It follows thatγ ≡ 1 or 2 (mod 3) and
γ ≡ 3 or 4 (mod 5). Thus the only possibilites forγ are 22 or 23 and so 3· 5 | β δ. If 5 | δ then we
take the equationγz4 − δw4 = 2 modulo 5 and we getγ ≡ 2 (mod 5). This cannot hold forγ = 22
or 23 so we must have 5∤ δ and so 5| β must occur. Takingβv4 − γz4 = 1 modulo 5 we get that
γ ≡ 4 (mod 5) and soγ = 22 is the only possibility. Becauseα = 1 andγ = 22 and 5| β, the only
possibilities forδ are 1, 3, or 6. However, taking the equationβv4 − δw4 = 3 modulo 5 we get that
δ ≡ 2 (mod 5). Therefore none of the possibilities work, and thusα , 1.
This exhausts all of the possibilities forα. Thus there can be no integral solutions for the curve
CS F(r, s). �
R
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wendung auf die Losung einer Klasse von bestimmter Gleichungen vienten Grades”,Det Norske Vidensk. Akad. Oslo
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