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Pythagorean Triples with a leg length being a power

of 2; or a power of an odd prime; or a perfect square

Konstantine Zelator

Department of Mathematics

317 Bishop Hall

SUNY Buffalo State College

1300 Elmwood Avenue

Buffalo, NY, 14222

Also, K. Zelator

P.O Box 4280

Pittsburgh, PA 15203 U.S.A

e-mail address: 1.) konstantine_zelator@yahoo.com

2.) zelatok@buffalostate.edu

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1. Introduction

As indicated by the title of this work, there are three types of Pythagorean

triples (or triangles) studied and investigated in this paper. First, the

Pythagorean triples that have a leg length, which is a power of 2. In Result 1

we describe all primitive Pythagorean triangles with one leg length being a

power of 2. After that, in Result 2, we give a parametric description of those

non-primitives Pythagorean triples with a leg length being a power of 2. Both

Results 1 and 2 are found in Section 4. In Section 5, we present Results 3 and

4, in which parametric descriptions are given for the family of primitive

Pythagorean triangles with one leg length being a power of an odd prime

(Result 3); and for the set of non-primitive Pythagorean triples with the same

property. In Section 2, we state the well-known parametric formulas that

describe the entire family of Pythagorean triples. In Section 3, the reader will

find Lemma 1 (Euclid’s lemma); Propositions 1, 2, and 3; and Theorem 1. All

five of these results are very well known in number theory. Lemma 1, and

Propositions 1, 2, and 3; are used in various proofs, especially in the proof of

Propositions 4 and 5 (found in Section 6)

Using Propositions 4 and 5, we establish Results 5 and 6 (in Section 7). In

Result 5, we give a parametric description of the set of primitive Pythagorean

triangles one of whose lengths is a perfect or integer square. In Result 6, we

present a parametric description of the family of non-primitive Pythagorean

triples that have a leg length, which is a perfect square. In Section 8, we have

Theorem 2. Theorem 2 postulates that there exists no Pythagorean triangle

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with one leg length being a perfect square; the other leg length being of the

form pk; where p is an odd prime such that p≡5 or 7(mod8); and k an odd

positive integer. Combining Theorem 2 with Theorem 1, we establish the

Corollary to Theorem 2. This article end in Section 9, which contains Theorem

3 and 4. Theorem 3 states that there exists no Pythagorean triple with one leg

length being a power of 2, the other leg length a perfect square. Finally,

according to Theorem 4, the only Pythagorean triple with one leg length being

a power of 2, the other a power of p (an odd prime); is the triple (3, 4, 5).

2. Pythagorean Triples

Let a, b, c be positive integers. The triple (a, b, c) is called a Pythagorean

triple or triangle with hypotenuse length c and leg lengths a and b; if

a2+b2=c2. The following formulas, very well-known, Formulas (1); describe

the entire family or set of Pythagorean triples. A derivation of Formulas (1)

can be found in multiple sources; for example see reference [1] or [2].

Reference [1] contains a wealth of information on Pythagorean triples. Also,

reference [3].

Formulas (1)

A triple (a, b, c) of positive integers a, b, c; is a Pythagorean

triple (with hypotenuse length c) if, and only if a=d(m2-n2),

b=d(2mn), c=d(m2+n2) (or alternatively, a=d(2mn),

b=d(m2-n2)) where d, m, n are positive integers such that

m>n, gcd(m, n)≡1 (mod2) (i.e one of m, n is odd, the other

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even) (Note: As it is easily shown, gcd(2mn,m2-n2)=gcd(2mn,

(m2+n2)=gcd((m2-n2, m2+n2)=1. If d=1, the Pythagorean

triple is called primitive. (1)

3. Five results from number theory

The first four of these results; Lemma 1, and Propositions 1 and 2; and can

easily be found in number theory books. For example, see references [1] and

[2].

Lemma 1 (Euclid’s lemma)

Let a, b, c be positive integers and suppose that a is a divisor of the product

bc. Furthermore, assume that gcd(a, b)=1 (i.e a and b are relatively prime).

Then, a|c; i.e the integer a must be a divisor of the integer c.

Proposition 1

Suppose that a, b, n are positive integers such that, an|bn. Then a|b.

Proposition 2

Let a, b, c be positive integers with a and b being relatively prime;

gcd(a,b)=1. Also, assume that, ab= c2. Then there exist relatively prime

positive integers d and e, such that a=d2, b=e2, and c=de.

The following proposition, Proposition 3; can be proved using Lemma 1 and

Proposition 2. We invite the reader to do so.

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Proposition 3

(i) Let p be a prime, and a, b, c positive integers such that ab=pc2. And

with a and b being relatively prime; gcd(a, b)=1. Then, there exist

relatively prime positive integers d and e such that, a=pd2, b=e2, and

c=de; or alternatively, a=d2, b=pe2, and c=de

(ii) Let p be a prime, and a, b, c ∈ ℤ+; and with gcd(a, b)=1 and pab=c2.

Then, there exist relatively prime integers d and e such that a=pd2,

b=e2, and c=pde; or alternatively, a=d2, b=pe2, and c=pde.

Theorem 1 below, is well known in the literature of number theory. It can be

found in some number books; for example see reference [1] (where the reader

can find a proof of this result.) Historically, this theorem is attributed to P.

Fermat.

Theorem 1

The 3-variable equation x4+y4=z2, has no positive integer solutions.

An immediate consequence of Theorem 1, is that there exists no Pythagorean

triangle with both leg lengths being perfect or integer squares.

4. Results 1 and 2

Result 1

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(i) There exists no primitive Pythagorean triple with a leg length equal to

2.

(ii) If k  ∈ ℤ+ and k≥2, there exists a unique primitive Pythagorean triple

with one leg length equal to 2k. That triple (a, b, c) is given by (may

switch a with b), a=22(k-1)-1, b=2k, c=22(k-1)+1

Proof

A primitive triple requires d=1 in formulas (1). We have,

a=m2-n2, b=2mn, c=m2+n2

m, n ∈ ℤ+, m > n. gcd(m, n)=1, m+n≡1(mod2) (2)

(i) Since mn≡0(mod2); 2mn≡0(mod4), and so it is clear that b cannot be

equal to 2.

(ii) Since b is the even leg length (a is the odd leg length) we must have,

b=2k=2mn; 2k-1=mn (3)

In (3) we we have k-1≥1, gcd(m, n) =1, m>n; and with one of m, n being odd,

the other even. These conditions clearly imply in (3); that we must have

m=2k-1 and n=1 (3a)

From (3a) and the formulas in (2), we obtain a=22(k-1)-1, b=2k, and c=22(k-1)+1

(3b).

Conversely, if k is a positive integer, k≥2; and the positive integers a, b, c

satisfy the formulas in (3b), then a straightforward calculation shows that

a2+b2=c2; and so (a, b, c) is a Pythagorean triangle with one leg length being

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2k. And it is primitive since, by inspection, we have gcd(a, b)=gcd(a, c)=gcd(b,

c)=1.

We now state Result 2

Result 2

(i) There is no non-primitive Pythagorean triangle with a leg length

equal to 2.

(ii) Let k ∈ ℤ+, k≥2. There are exactly (k-1) non-primitive Pythagorean

triples with a leg length being 2k. These (k-1) triples are given by the

parametric (one-parameter) formulas,

a=2t[22(k-1-t)-1], b=2k, c=2t[22(k-1-t)+1]; where t=1,….,k-1

Proof

Let (a, b, c) be a non-primitive Pythagorean triple. Then, according to Formulas

(1), we must have,

a=d(m2-n2), b=d(2mn), c=d(m2+n2);

with d, m, n ∈ ℤ+, d≥2, m>n,

gcd(m, n)=1, and m+n ≡1(mod2) (4)

First, observe that the positive integer a, cannot equal 2k. Indeed, if it were

2k=d(m2-n2); then since m2-n2 is odd (in virtue of the fact that one of m, n is odd;

the other even); Lemma 1 would imply that 2k|d; d=(2k)(D), for some D ∈ ℤ+.

Thus, 1=D(m2-n2), which obviously is impossible since m2-n2=(m-n)(m+n)≥3, on

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account of the fact that m, n ∈ ℤ+ and m>n; and so m-n≥1 and m+n≥3 (due to

the fact that m and n have different parities). It is now clear that if the triple (a, b,

c) has a leg length equal to 2k. Then b=d(2mn) must be that length. We have,

b=d(2mn)=2k or dmn=2k-1 (4a)

Since mn≡0 (mod2); equation (4a) makes it clear that k≥2. This establishes part

(i). For part (ii): By (4), it is clear that

d=2t; 0≤t≤k-1

t∈ ℤ (4b)

But d≥2, since (a, b, c) is a non-primitive triple. Hence, t cannot equal zero;

which means that by (4b)

d=2t, t=1,…,k-1

k≥2, k ∈ ℤ (4c)

Combining (4a) with (4c) yields

mn=2k-1-t (4d)

The conditions m>n, gcd(m, n)=1, and m+n≡1(mod2) in (4); together with

equation (4d) imply that

m=2k-1-t and n=1 (4e)

Combining (4e), (4c), and (4) yields,

a=2t[22(k-1-t)-1], b=2k, c=2t[22(k-1-t)+1] (4f)

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Conversely, if a positive integer triple (a, b, c) satisfies the formulas in (4f); a

routine calculation shows that a2+b2=c2; then showing that (a, b, c) is a

Pythagorean triple; clearly non-primitive with d=2t; since 1≤t≤k-1.

5. Results 3 and 4

Let p be an odd prime and k a positive integer. If (m2-n2, 2mn, m2+n2) is a

primitive Pythagorean triple (Formulas (1) with d=1), the odd leg length is m2-n2.

Thus, if a primitive triple has a leg length pk; we must have

m2-n2=pk; (m+n)(m-n)=pk (5)

Since gcd(m, n)=1, m > n, and m+n≡1(mod2); it follows that

gcd(m+n, m-n)=1

Also, m+n>m-n≥1 (5a)

From (5) and (5a) if follows that,

m+n=pk and m-n=1;

m=!!!!!

and n=!!!!!

(5b)

Combining formulas (5b) with Formulas (1), with d=1; we arrive at Result 3.

Result 3

Let p be an odd prime and k a positive integer. There exists a unique primitive

Pythagorean triple (a, b, c) with one of its leg lengths being pk; the triple,

a=pk, b=2(!!!!!

)(  !!!!!

), c=!!!!!!

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b=!!!!!!

Next, consider a non-primitive Pythagorean triangle with one leg length of the

form pk. A non-primitive triple has the form (d(m2-n2), d(2mn), d(m+n)), with

d≥2; and the positive integers m and n satisfying the conditions in Formulas (1).

Since d(2mn) is even, it cannot equal pk. Thus we must have,

d(m+n)(m-n)=pk (5c)

Obviously, since d≥2, (5c) shows that case k=1 is not possible; since in such a

case (5c) would imply that (m+n)(m-n)≥3 (in virtue of m, n 𝜖 ℤ+, m>n, and

m+n≡1(mod2)).

Therefore, given that d≥2; (5c) implies that k≥2 and

d=pt, m+n=pk-t, m-n=1;

m=!!!!!!

! , n=

!!!!!!!

, d=pt; t=1,….k-1 (5d)

A calculation shows that d(2mn)=b=(!!)(!! !!! !!)

!,

pk=a=d(m2-n2), and c=!(!!!!!)

!=!!(!! !!! !!)

!

We can now state Result 4.

Result 4

Let p be an odd prime

(i) There exists no non-primitive Pythagorean triple having a leg length

equal to p.

(ii) Let k  𝜖  ℤ+, k≥2.

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There exist exactly (k-1) non-primitive Pythagorean triangles (a, b, c), with

one leg length being pk. These (k-1) triples are given by,

a=pk, b=!!(!! !!! !!)!

, and c=!!(!! !!! !!)!

and with t=1,….,k-1

6. Propositions 4 and 5

Proposition 4

Consider the four-variable equation xy=zw2, under the co-primeness

condition gcd(x, y)=1. Then, all the positive integer solutions to this equation

can be described in terms of four parameters (𝛿, T, R, and u):

x=𝛿R2, y=uT2, w=TR, z=𝛿u

where 𝛿, T, R and u can be any positive integers such

that gcd(𝛿R, uT)=1 (Formulas (6))

Proof

We have,

xy=zw2

gcd(x, y)=1 (6a)

A straightforward calculation shows that is x, y, z, w satisfy Formulas (6);

then the equation in (6a) is satisfied. Also, the condition gcd(𝛿R, uT)=1,

clearly implies that gcd(uT2, 𝛿R2)=1=gcd(x, y). Now the converse. Assume

that x, y, z, w 𝜀  ℤ+; satisfy (6a).

Let 𝛿=gcd(x, z), 𝛿  the greatest common divisor of x and z. Then,

x=𝛿v, z=𝛿u;

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where u and v are relatively prime positive integers; (6b)

gcd(v, u)=1

From (6b) and (6a) we obtain

𝛿vy=𝛿uw2;

vy=uw2 (6c)

Since gcd(v, u)=1 and u|vy in (6c); Lemma 1 (Euclid’s lemma) implies that

u|y:

y=ut (6d), t 𝜀 ℤ+

Combining (6d) with (6c) gives,

vt=w2 (6e)

Since v is a divisor of x; and t is a divisor of y; and gcd(x, y)=1. It follows that,

gcd(v, t)=1 (6f)

Proposition 2, (6e), and (6f) imply that

t=T2, v=R2, w=(T)(R);

with T, R relatively prime positive integers;

gcd(T, R)=1 (6g)

Going back (6d) and (6b), we have

x=𝛿(R2), y=u(T2), z=𝛿u, w=(T)(R)

with gcd(T, R)=1. But also, by (6a) the condition gcd(x, y)=1; requires that in

fact gcd(𝛿R2, uT2)=1. Which is equivalent to gcd(𝛿R, uT)=1; or equivalently

gcd(𝛿, T)=gcd(R, u)=gcd(R, T)=1. The proof is complete.

We now use Proposition 4, in order to establish Proposition 5.

Proposition 5

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Consider the 4-variable equation xyz=w2, under the co-primeness condition

gcd(x, y)=1. Then all the positive integer solutions to this equation can be

described in terms of five parameters (𝛿, R, T, u and Z):

x=(𝛿)(R2), y=(u)(T2), z=𝛿(u)(Z2), w=(𝛿)(u)(R)(T)(Z)

where 𝛿, T, R, u, and Z can be any positive integers such that gcd((𝛿)(R),

(u)(T))=1.

Proof

We have,

xyz=w2

gcd(x, y)=1 (7a)

A straightforward calculation show that if x, y, z, w satisfy Formulas (7); then

the equation (7a) is satisfied. Also, the condition gcd((𝛿)(R), (u)(T))=1 clearly

implies that gcd((𝛿)(R2), (u)(T2))=1=gcd(x, y). Now the converse statement.

Assume that x, y, z, w 𝜖  ℤ+ satisfy (7a). Let g=gcd(z, w); so that

z=(g)(Z), w=(g)(W),

for positive integers Z, W; such that gcd(Z, W)=1 (7b)

From (7a) and (7b) we obtain,

xygZ=g2W2;

xyZ=gW2 (7c)

Since gcd(Z, W)=1, we have gcd(Z, W2)=1. Thus by (7c) and Lemma 1, it

follows that Z|g:

g=(Z)(V) (7d)

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Combining (7c) and (7d), we get

xy=(V)(W2)

And with gcd(x, y)=1 (by (7a) ) (7e)

Proposition 4 together with (7e) imply that

x=(𝛿)(R2), y=(u)(T2), v=(𝛿)(u), W=(R)(T),

for positive integers 𝛿, R, T, u; such that gcd(𝛿R, uT)=1) (7f)

Using (7f), (7d), and (7b); we also obtain z=(𝛿)(u)(Z2) and

w=(𝛿)(u)(R)(T)(Z). The proof is complete.

7. Results 5 and 6

Consider a primitive Pythagorean triple (m2-n2), 2mn, m2+n2)’ with one leg

length being a perfect square. If the odd length is a perfect square; then

m2-n2=k2; m2=n2+k2 (8)

Since gcd(m, n)=1 and m+n≡1(mod2). It follows from (8), that (k, n, m) is

actually a primitive Pythagorean triple with hypotenuse length m. Clearly m

must be odd and n even (the hypotenuse of a primitive triple is always odd;

according to Formulas (1), with d=1. Therefore, we must have

m=M2+N2, n=2MN, k=M2-N2

where M, N are positive integers such that M>N,

gcd(M, N)=1 and M+N≡1(mod2) (8a)

The other possibility; of course, is that the even leg length in a primitive triple

is a perfect square:

2mn=K2 (9)

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We apply Proposition 3 in (9), with p=2: we must have either

m=2r2, n=s2; with r, s 𝜖  ℤ+; s odd, and

gcd(r, s)=1 and with 2r2>s2 (since m>n) (9a)

Or alternatively,

m=r2, n=2s2;

with r, s 𝜖  ℤ+; r odd, an gcd(r, s)=1;

and with r2>2s2 (9b)

Having (8a), (9a), and (9b) in place; we can now formulate Result 5.

Result 5

The set of all primitive Pythagorean triples (m2-n2, 2mn, m2+n2), with one

leg length being a perfect square; is the union of three families of primitive

Pythagorean triples

Family F1: (m2-n2, 2mn, m2+n2); with m=M2+N2, n=2MN; where M, N are

positive integers such that M>N gcd(M, N)=1 and M+N≡1(mod2). And so

(m2-n2, 2mn, m2+n2)=((M2-N2)2, 4MN(M2-N2), M4+6M2N2+N4)

Family F2: (m2-n2, 2mn, m2+n2); with m=2r2, n=s2; r, s 𝜖  ℤ+, gcd(r, s)=1

and s≡1(mod2); and with 2r2>s2 And so, (m2-n2, 2mn, m2+n2)=(4r4-s4, 4r2s2,

4r4+s4)

Family F3: (m2-n2, 2mn, m2+n2); with m=r2, n=2s2; r, s 𝜖  ℤ+, gcd(r, s)=1

and r≡1(mod2); and with r2>2s2 And so, (m2-n2, 2mn, m2+n2)=(r4-4s4, 4r2s2,

r4+4s4)

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Next we establish Result 6. Consider a non-primative Pythagorean triangle with a

leg length a perfect square. The two legs together are

𝑑(𝑚 + 𝑛)(𝑚 − 𝑛) and 𝑑  (2𝑚𝑛)

Suppose first that 𝑑 𝑚 + 𝑛 𝑚 − 𝑛 = 𝐾!, with 𝑑 ≥ 2 (since the triangle is

nonprimitive).

From Formula (1) we know that gcd 𝑚,𝑛 = 1,𝑚 > 𝑛,𝑚 + 𝑛 ≡ 1 𝑚𝑜𝑑2 .

Therefore, gcd 𝑚 + 𝑛,𝑚 − 𝑛 = 1;𝑚 + 𝑛 and 𝑚 − 𝑛are relatively prime odd

positive integers. We have,

𝑑 𝑚 + 𝑛 𝑚 − 𝑛 = 𝐾! , gcd 𝑚 + 𝑛,𝑚 − 𝑛 = 1,𝑑 ≥ 2) (9c)

We apply Proposition5 to (9c):

𝑚 + 𝑛 = 𝛿 ∗ 𝑅!,𝑚 − 𝑛 = 𝑢 ∗ 𝑇!,𝑑 = 𝛿 ∗ 𝑢 ∗ 𝑍!,𝐾 = 𝜕 ∗ 𝑢 ∗ 𝑅 ∗ 𝑡 ∗ 𝑍   𝟗𝐝

With 𝛿, R, u, T, Z being positive integers such that 𝛿 ≡ 𝑅 ≡ 𝑢 ≡ 𝑇 ≡ 1(𝑚𝑜𝑑2) (so

d, R, u, T are all odd), and with gcd 𝛿 ∗ 𝑅,𝑢 ∗ 𝑇 = 1; also with at least one 𝛿,u Z

being greater than 1 (𝑠𝑜  𝑡ℎ𝑎𝑡  𝑑 ≥ 2).

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We obtain, from (9d),

𝑚 = !!!!!!

!,𝑛 = !!!!!!

!, so we must also have 𝛿𝑅! > 𝑢𝑇!; since 𝑛 ≥ 1.

A computation shows that the other leg lengths are,

𝑑 2𝑚𝑛 = 𝛿 ∗ 𝑢 ∗ 𝑍! ∗𝑑!𝑅! − 𝑢!!!

2 ,𝑎𝑛𝑑

𝑑 𝑚! + 𝑛! = 𝛿 ∗ 𝑢 ∗ 𝑍! !!!!!!!!

!

!.

Now, lets look at the case in which the other leg length, 𝑑 2𝑚𝑛 , is a perfect

square. We have,

𝑑 2𝑚𝑛 = 𝐿!;𝑎𝑛𝑑  𝑠𝑖𝑛𝑐𝑒  𝐿  𝑖𝑠  𝑒𝑣𝑒𝑛; 𝐿 = 2𝐹;𝐹 ∈ ℤ!.𝑑 2𝑑𝑚𝑛 = 4𝐹!(9e)

We know that either m is even and n is odd; or vice-versa. If m is even and n odd;

𝑚 = 2𝑀.

Then by (9e),

𝑑 ∗𝑀 ∗ 𝑛 = 𝐹! (9f)

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Since,

gcd 𝑚,𝑛 = 1 = gcd 2𝑀,𝑛 = 1;𝑤𝑒  ℎ𝑎𝑣𝑒 gcd 𝑀,𝑛 = 1. (9g).

Proposition 5, (9f), and (9g) imply that,

𝑀 = 𝑑𝛿; and so 𝑚 = 2𝛿𝑅!.

And 𝑛 = 𝑢 ∗ 𝑇!, 𝑑 = 𝛿 ∗ 𝑢 ∗ 𝑍!,𝐹 = 𝛿 ∗ 𝑢 ∗ 𝑅 ∗ 𝑇 ∗ 𝑍. (9h)

With 𝑑,𝑅,𝑢,𝑇 being positive integers such that

𝑢 ≡ 𝑇 ≡ 1 𝑚𝑜𝑑  2 , gcd 𝛿𝑅,𝑢𝑇 = 1,𝑎𝑛𝑑  2𝛿𝑅! > 𝑢𝑇!, 𝑠𝑖𝑛𝑐𝑒  𝑚 > 𝑛. And with

atleast one of 𝛿, u, Z being greater than 1; since 𝑑 ≥ 2. So the other two leg

lengths are 𝑑 𝑚! − 𝑛! = 𝛿 ∗ 𝑢 ∗ 𝑍! ∗ 4𝛿!𝑅! − 𝑢!𝑇! ;𝑎𝑛𝑑  𝑑 𝑚! + 𝑛! = 𝛿 ∗ 𝑢 ∗

𝑍!(4𝛿  𝑅! + 𝑢!𝑇!)

The other case in (9e) is m odd and n even (𝑛 = 2𝑁) and it is treated similarly.

We can now state Result 6.

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Result 6

The set of non-primitive Pythagorean triangles with a leg length being a perfect

square; is the union of the following three families on non-primitive.

Pythagorean triples 𝑎, 𝑏, 𝑐 :

Family 𝑭𝟏:  𝑎 = 𝛿 ∗ 𝑢 ∗ 𝑅 ∗ 𝑇 ∗ 𝑍 !, 𝑏 = 𝛿 ∗ 𝑢 ∗ 𝑍! ∗ !!!!!!!!!

!,𝑎𝑛𝑑  𝑐 = 𝛿 ∗ 𝑢 ∗

𝑍! !!!!!!!!

!;  where 𝛿,𝑢,𝑅,𝑇,𝑍 are positive integers such that

𝛿𝑅! > 𝑢𝑇!,𝑔𝑐𝑑 𝛿𝑅,𝑢𝑇 = 1. And with all four 𝛿,𝑢,𝑅,𝑇 being odd: 𝛿𝑢𝑅𝑇 ≡

1 𝑚𝑜𝑑2 . Also, with at least one of 𝛿, u, Z greater than 1 so that 𝛿 ∗ 𝑢 ∗ 𝑍! ≥ 2.

Result 6

𝑭𝒂𝒎𝒊𝒍𝒚  𝑭𝟐:𝑎 = 𝛿 ∗ 𝑢 ∗ 𝑍! ∗ 4𝛿!𝑅! − 𝑢!𝑇! , 𝑏 = 2𝛿𝑢𝑅𝑇𝑍 !, 𝑐

= 𝛿 ∗ 𝑢 ∗ 𝑍! 4𝛿!𝑅! + 𝑢!𝑇!

with 𝛿,𝑢,𝑅,𝑇,𝑍  being positive integers such that 2𝛿𝑅! > 𝑢𝑇!,𝑔𝑐𝑑 𝛿𝑅,𝑢𝑇 = 1;

and with u, T being odd: 𝑢𝑇 ≡ 1 𝑚𝑜𝑑2 .    And also with at least one of 𝛿, u, Z

being greater than 1; so that 𝛿 ∗ 𝑢 ∗ 𝑍! ≥ 2.

𝑭𝒂𝒎𝒊𝒍𝒚  𝑭𝟑:𝑎 = 𝛿 ∗ 𝑈 ∗ 𝑍! 𝛿!𝑅! − 4𝑢!𝑇! ,  

𝑏 = 2𝛿𝑢𝑅𝑇𝑍 !,

𝑐 = 𝛿𝑢𝑍! 𝛿!𝑅! + 4𝑢!𝑇! :

Where 𝛿, u R, T, Z are positive integers such that 𝑑𝛿 > 2𝑢𝑇!, gcd 𝛿𝑅,𝑢𝑇 =

1, 𝛿  𝑎𝑛𝑑  𝑅  𝑏𝑜𝑡ℎ  𝑜𝑑𝑑, 𝛿𝑅 ≡ 1 𝑚𝑜𝑑2 .

And with atleast one of 𝛿, u, Z greater than 1; so that 𝛿 ∗ 𝑢 ∗ 𝑍! ≥ 2.

  20  

8. Theorem 2 and its proof

Theorem 2

Let p be an odd prime such that 𝑝 ≡ 5𝑜𝑟  7 𝑚𝑜𝑑 ∗ 8 ; and k an odd positive

integer. There exists no Pythagorean triangle with one leg length being 𝑝! and

the other leg length a perfect square.

Proof

The two leg lengths, according to Formula (1), are 𝑑 𝑚! − 𝑛! 𝑎𝑛𝑑  𝑑 2𝑚𝑛 . First

observe that 𝑑 2𝑚𝑛 cannot obviously equal 𝑝! , since 𝑑(2𝑚𝑛) is even and 𝑝! is

odd. Therefore, if such Pythagorean triple exists, then we must have

𝑝! = 𝑑 𝑚 + 𝑛 𝑚 − 𝑛 𝑎𝑛𝑑  𝑑 2𝑚𝑛 = 𝐿!,

𝑤ℎ𝑒𝑟𝑒  𝑑,𝑚,𝑛, 𝐿  𝑎𝑟𝑒  𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒  𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠, 𝑠𝑢𝑐ℎ  𝑡ℎ𝑎𝑡  𝑚 > 𝑛,𝑔𝑐𝑑 𝑚,𝑛 = 1,𝑚 + 𝑛 ≡

1 𝑚𝑜𝑑2 (one of m, n is odd, the other even); and with obviously L even. (10)

Clearly the first equation in (10), shows that d must be a non-negative integer

power of 𝑝;𝑑 = 𝑝!;𝑤𝑖𝑡ℎ  0 ≤ 𝑡 ≤ 𝑘. But t cannot equal k; for this would imply that

𝑔𝑐𝑑 𝑚 + 𝑛 , (𝑚 − 𝑛) = 1,𝑛𝑜𝑡  𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒, 𝑠𝑖𝑛𝑐𝑒   𝑀 + 𝑛 𝑚 − 𝑛 ≥ 3, according to the

conditions than m, n satisfy in (10). We have,

𝑑 = 𝑝! , 0 ≤ 𝑡 < 𝑘; 0 ≤ 𝑡 ≤ 𝑘 − 1 (10a)

From (10a) and (10) we obtain,

𝑝!!! = (𝑚 + 𝑛)(𝑚 − 𝑛) (10b)

  21  

Due to the conditions in (10), the odd positive integers m+n and m-n are

relatively prime; and since obviously 𝑚 + 𝑛 > 𝑚 − 𝑛 ≥ 1; equation (10b) implies

𝑚 + 𝑛 = 𝑝!!!𝑎𝑛𝑑  𝑚 − 𝑛 = 1. There fore, (2𝑚 = 𝑝!!! + 1  𝑎𝑛𝑑  2𝑛 = 𝑝!!! − 𝟏 (10c)

Clearly, (10c) implies that neither m nor n is divisible by 𝑝:

𝑚𝑛 ≢ 0(𝑚𝑜𝑑𝑝) (10d)

From (10a) and (10) we also have,

𝑝! ∗ 2𝑚𝑛 = 𝐿! (10e)

Since mn is not divisible by p; and since the highest power of p dividing 𝐿!, must

be even (including the possibility of zero exponent); it follows that since 𝑝! is the

highest power of p dividing the left-hand side of (10e); the exponent t must

being even:

𝑡 = 2𝑟, 𝑟 is non-negative integer (10f)

Thus, 𝑝!!  𝑑𝑖𝑣𝑖𝑑𝑒𝑠  𝐿!𝑖𝑛 (10e) which implies by Proposition 1 that,

 𝑝!  𝑑𝑖𝑣𝑖𝑑𝑒𝑠 ∶ 𝐿 = 𝑝! ∗ 𝑙  (10g)

  22  

where  𝑙 is a positive integer.

From (10g), (10f), and (10e) we get,

2𝑚𝑛 = 𝑙! (10h)

Proposition 3 (for p=2), the fact that gcd 𝑚,𝑛 = 1; and (10h); imply that either

Possibility 1: 𝑚 = 2𝑀!  𝑎𝑛𝑑  𝑛 = 𝑁!; with M and N being positive integers, with

N odd; 𝑁 ≡ 1(𝑚𝑜𝑑2) (since m and n have different parities; one is even, the other

odd) And with gcd 𝑀,𝑁 = 1. Or

Possibility 2: 𝑚 = 𝑀!,𝑎𝑛𝑑  𝑛 = 2𝑁!:𝑀,𝑁   ∈ ℤ!,𝑀  𝑜𝑑𝑑;𝑎𝑛𝑑 gcd 𝑀,𝑁 = 1.

Suppose Possibility 1 holds true. We go back to (10c), with 𝑡 = 2𝑟 (from (10f)).

We have 4𝑀! = 𝑝!!!! + 1  𝑎𝑛𝑑  2𝑁! = 𝑝!!!! − 1 (10i). Since k is odd; the

positive integer k-2r is odd: 𝑘 − 2𝑟 = 2𝑣 + 1; v is a non-negative integer.

Thus (10i) gives, 4𝑀! = 𝑝!!!! + 1  𝑎𝑛𝑑  2𝑁! = 𝑝!!!! − 1. Modulo 8, (10i) is

impossible if 𝑝 ≡ 5  𝑜𝑟  7(𝑚𝑜𝑑8). Indeed if 𝑝 ≡ 5 𝑚𝑜𝑑8 ;  𝑡ℎ𝑒𝑛    𝑝!!!! ≡ 𝑝! ! ∗

𝑝 ≡ 1 ∗ 𝑝 ≡ 5 𝑚𝑜𝑑8  𝑠𝑖𝑛𝑐𝑒  𝑝! ≡ 1(𝑚𝑜𝑑8) (the square of any odd integer is

congruent to 1 modulo 8).

  23  

But then 𝑝 !!!! − 1 ≡ 5− 1 ≡ 4 𝑚𝑜𝑑8 ,𝑤ℎ𝑖𝑙𝑒  2𝑁! ≡ 2 ∗ 1 ≡ 2 𝑚𝑜𝑑8 ;  whick

renders the second equation in (10i) contradictory.

Now, if 𝑝 ≡ 7 𝑚𝑜𝑑8 ; 𝑡ℎ𝑒𝑛  𝑝!!!! − 1 ≡ 1 ∗ 7− 1 ≡ 7− 1 ≡ 6 𝑚𝑜𝑑8 .  Again

rendering the second equation in (10i) contradictory. Next, if Possibility 2 holds

true. Then by (10c) with 𝑡 = 2𝑟 we have,

2𝑀! = 𝑝!!!! + 1  𝑎𝑛𝑑

4𝑁! = 𝑝!!!! − 1; 2𝑀! = 𝑝!!!! + 1 and

4𝑁! = 𝑝!!!! − 1 (10j)

The first equation in (10j) is impossible when 𝑝 ≡ 5  𝑜𝑟  7 𝑚𝑜𝑑8 .    𝑖𝑛𝑑𝑒𝑒𝑑  2𝑀! ≡

2 𝑚𝑜𝑑8 ,  since M is odd. But

𝑝!!!! ≡ 𝑝 + 1 ≡ 6  𝑜𝑟  0 𝑚𝑜𝑑8 ; 𝑓𝑜𝑟  𝑝 ≡ 5  𝑜𝑓  7 𝑚𝑜𝑑8 ,  respectively. The proof is

complete. ∎

From Theorems 1 and 2, we have the following immediate corollary (since by Th.

1, k cannot be even)

Corollary: Let k be a positive integer, p a prime with 𝑝 ≡ 5  𝑜𝑟  7 𝑚𝑜𝑑8 . There

exists no Pythagorean triangle with one leg length 𝑝! ,  the other a perfect

square.

  24  

9. Theorems 3 and 4

Theorem 3

Let k be a positive integer. There exists no Pythagorean triangle with one leg

length equal to 2!, the other leg length being a percfect square.

Proof

First observe that if such a triangle exists, the leg length 𝑑 𝑚! − 𝑛! cannot equal

2!; 𝑠𝑖𝑛𝑐𝑒  𝑚! − 𝑛! = 𝑚 + 𝑛 𝑚 − 𝑛 ≥ 3;𝑎𝑛𝑑  𝑠𝑜  𝑚! − 𝑛! must have an odd prime

divisor; in view of the fact that 𝑚! − 𝑛! is odd. Therefore, if such s triangle exists;

we must have,

𝑑 𝑚 + 𝑛 𝑚 − 𝑛 = 𝐿!,𝑑 2𝑚𝑛 = 2!;

𝑤ℎ𝑒𝑟𝑒  𝑑,𝑚,𝑛, 𝐿  𝑎𝑟𝑒  𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒  𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠  𝑠𝑢𝑐ℎ  𝑡ℎ𝑎𝑡  𝑚 > 𝑛, gcd 𝑚,𝑛 = 1  𝑎𝑛𝑑  𝑚 +

𝑛 ≡ 1(𝑚𝑜𝑑2) (11)

Note that since 𝑚𝑛 ≡ 0 𝑚𝑜𝑑2 ;

2𝑚𝑛 ≡ 0 𝑚𝑜𝑑4 ; the positive integer k must be at least 2: 𝑑 𝑚𝑛 = 2!!!, 𝑘 ≥ 2

(11a)

  25  

Obviously d must be a power of 2; ad since mn in even; (11a) implies that,

𝑑 = 2! , 𝑡  𝑎  𝑛𝑜𝑛 − 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒  𝑖𝑛𝑡𝑒𝑔𝑒𝑟  𝑠𝑢𝑐ℎ  𝑡ℎ𝑎𝑡  0 ≤ 𝑡 ≤ 𝑘 − 2 (11b).

In fact t must be even, since according to the first equation in (11): the highest

power dividing the left-hand side is 2! , 𝑠𝑖𝑛𝑐𝑒   𝑚 + 𝑛 𝑚 − 𝑛 is odd. But the

right-hand side is 𝐿!; which shows that t must be even: 𝑡 = 2𝑟   𝟏𝟏𝒄 .

Thus, since 𝑑 = 2!!𝑑𝑖𝑣𝑖𝑑𝑒𝑠  𝐿!; it follows by Proposition 1 that 2!𝑑𝑖𝑣𝑖𝑑𝑒𝑠  𝐿; and so

𝐿 = 2! ∗ 𝑙, 𝑙 ∈ ℤ! (11d).

From (11d), (11a), and (11) we obtain [ 𝑚 + 𝑛 𝑚 − 𝑛 = 𝑙!,𝑚𝑛 = 2!!!!!](11e).

Since gcd 𝑚,𝑛 = 1,𝑎𝑛𝑑  𝑚 > 𝑛. It follows from (11e) that,

𝑚 = 2!!!!!𝑎𝑛𝑑  𝑛 = 1 (11f).

Therefore from (11f) and (11e) we have,

𝑚! − 𝑛! = 𝑙!; 2! !!!!! − 1 = 𝑙!; ; 2! !!!!! = 1+ 𝑙!     (11g).

  26  

However, (11g) is impossible modulo 4: since 𝑘 − 1− 𝑡 ≥ 1 (by (11b)), the left-

hand side of (11g) is congruent to zero modulo 4. But the right-hand since,

𝑙! + 1 ≡ 1  𝑜𝑟  2 𝑚𝑜𝑑4 ; according to whether 𝑙 is even or odd. The proof is

complete. ∎

Theorem 4

Suppose that k and e are positive integers; p an odd prime. The only

Pythagorean triple that has one leg length equal to 2! ,  and the other leg length

𝑝!; 𝑖𝑠  𝑡ℎ𝑒  𝑡𝑟𝑖𝑝𝑙𝑒   3, 4, 5 (𝑠𝑜  𝑘 = 2,𝑝 = 3, 𝑒 = 1).

Proof

Such a triple must satisfy,

𝑝! = 𝑑 𝑚 + 𝑛 𝑚 − 𝑛  𝑎𝑛𝑑  2! = 𝑑 2𝑚𝑛 ;𝑑,𝑚,𝑛,∈ ℤ!  ,𝑚 > 𝑛, gcd 𝑚,𝑛 = 1,𝑚 +

𝑛 ≡ 1(𝑚𝑜𝑑2) (12).

According to (12); d is in fact the greatest common divisor of 𝑝!𝑎𝑛𝑑  2!; and

obviously 𝑑 = 1. Consequently, because of the conditions in (12); one easily sees

that (12) further implies 𝑚 = 2!!!,𝑛 = 1;𝑎𝑛𝑑  𝑘 ≥ 2.    And also, 𝑚 + 𝑛 =

𝑝!𝑎𝑛𝑑  𝑚 − 𝑛 = 1; 𝑡ℎ𝑢𝑠  𝑚 = 1+ 𝑛 = 1+ 1 = 2.   Therefore 𝑝! = 2+ 1 = 3⟹ 𝑝 =

3  𝑎𝑛𝑑  𝑒 = 1.    We are done. ∎

  27  

References

[1] W. Sierpinnksi, Elementary Theory of Numbers, 480pp., WARSAW, 1964

ISBN:0-598-52758-3 For Pythagorean triangle, see pages 38-52

[2] David M. Burton, Elementary Theory of Numbers, 436pp, 2011, Seventh

Edition. The McGraw Hill Companies ISBN:978-0-07-338134-9. For

Pythagorean triples see pages 245-251

[3] http://en.wikipedia.org/wiki/pythagorean-triple