## David Burton Elementary Number Theory Solutions Chapter 2 Divisibility Theory In The Integers

**Page 15 Problem 1 Answer**

Here, we have to prove that a number is triangular if and only if it is of the form n(n+1)/2 for some n ≥ 1. It’s given that each of the numbers 1 = 1, 3 = 1 + 2, 6 = 1 + 2 + 3, 10 = 1 + 2 + 3 + 4,… represents the number of dots that can be arranged evenly in an equilateral triangle:

**Proof:** Remember that we proved in sec 1.1 that, 1 + 2 + 3 + …. + n = \(\frac{n(n+1)}{2}\). So, if f is triangular, then for some n, 1 + 2 + 3 + …. + n = f by definition, and so f = \(\frac{n(n+1)}{2}\) for some n, then

f = 1 + 2 + 3 + … + n

Thus, we proved that a number is triangular if and only if it is of the form n(n+1)/2 for some n ≥ 1

Here, we have to prove that the integer n is a triangular number if and only if 8n + 1 is a perfect square. it’s given that each of the numbers 1 = 1, 3 = 1 + 2, 6 = 1 + 2 + 3, 10 = 1 + 2 + 3 + 4,… represents the number of dots that can be arranged evenly in an equilateral triangle:

**Proof:** (⇒) if n is triangular, then there is a k ∃ n = \(\frac{k(k+1)}{2}\).

∴ 8n = 4k(k+1)

8n + 1 = 4k(k+1) + 1

= 4k^{2} + 4k + 1

= (2k+1)^{2}

Therefore n is triangular ⇒ 8n + 1 is a perfect square.

if 8n + 1 is a perfect square, then there is an integer k ∃ k^{2} = 8n + 1. Note that 8n+1 must be odd. k^{2} is odd, and so k is odd, so there is an r ∃ 2r + 1 = k

(2r+1)^{2} = 8n + 1

4r^{2} + 4r + 1 = 8n + 1

4r(r+1) = 8n

\(\frac{r(r+1)}{2}\) = n

Therefore, 8n + 1 is a perfect square ⇒ n is triangular.

Thus, we proved that integer n is a triangular number if and only if 8n+1 is a perfect square.

Here, we have to prove that the sum of any two consecutive triangular numbers is a perfect square. It’s given that each of the numbers, 1 = 1,3 = 1 + 2,6 = 1 + 2 + 3,10 = 1 + 2 + 3 + 4,… represents the number of dots that can be arranged evenly in an equilateral triangle:

**Proof:** We want to prove that if k and l are consecutive triangular numbers, then k+l is a perfect square.

Let, 1 + 2 + 3 + ⋯ + n = k, then 1 + 2 + 3 + ⋯ + n + n + 1 = l

∴ k + l = \(\frac{n(n+1)}{2}+\frac{n(n+1)}{2}+(n+1)\)

= n(n+1) = (n+1)

= (n+1)(n+1)

So, k + 1 is a perfect square.

Thus, we proved that the sum of any two consecutive triangular numbers is a perfect square.

Here, we have to prove that if n is a triangular number, then so are 9n + 1,25n + 3, and 49n + 6.

It’s given that each of the numbers, 1 = 1,3 = 1 + 2,6 = 1 + 2 + 3,10 = 1 + 2 + 3 + 4,… represents the number of dots that can be arranged evenly in an equilateral triangle:

**Proof :** Let 1 + 2 + 3 + ⋯ + k = n, then

9n + 1 = \(\frac{9 k(k+1)}{2}+1=\frac{\left.9 k^2+9 k+2\right)}{2}=\frac{(3 k+1)(3 k+2)}{2}=\frac{s(s+1)}{2}^{\text {for } s=3 k+1}\) 9n + 1 is a triangular.

25n + 3 = \(\frac{25 k(k+1)}{2}+3=\frac{\left.25 k^2+25 k+6\right)}{2}=\frac{(5 k+2)(5 k+3)}{2}=\frac{s(s+1)}{2}\) for s = 5k+2, thus 25n + 3 is triangular

49n + 6 = \(\frac{49 k(k+1)}{2}+6=\frac{\left.49 k^2+49 k+12\right)}{2}=\frac{(7 k+3)(7 k+4)}{2}=\frac{s(s+1)}{2}\) for s = 7k + 2, thus 49n + 6 is triangular.

Thus, proved that if n is a triangular number, then so are 9n + 1, 25n + 3, and 49n + 6.

**Page 15 Problem 3 Answer**

Here, we have to derive the following formula for the sum of triangular numbers, attributed to the Hindu mathematician Aryabhata,

t_{1}+ t_{2} + t_{3} + …. + \(\frac{n(n+1)(n+2)}{6}\) n ≥ 1. It’s given that, t_{k-1} + t_{k} = k^{2}

**Proof:** Recall from ex 1 (c) in sec 1.1,

1.2 + 2.3 + …. + n.(n+1) = \(\frac{n(n+1)(n+2)}{3}\), n ≥ 1

∴ \(\frac{1 \cdot 2}{2}+\frac{2 \cdot 3}{2}+\cdots+\frac{n \cdot(n+1)}{2}=\frac{n(n+1)(n+2)}{6}\)

Note that each term k can be written as \(\frac{k(k+1)}{2}=t_k\)

∴ t_{1} + t_{2} + t_{3} + …. + \(\frac{n(n+1)(n+2)}{6}\)

Thus, we derive the following formula for the sum of triangular numbers

t_{1} + t_{2} + t_{3} + …. + \(\frac{n(n+1)(n+2)}{6}\) n ≥ 1.

**Page 16 Problem 4 Answer**

Here, we have to prove that the square of any odd multiple of 3 is the difference of two triangular numbers specifically, that 9(2n+1)^{2} = t_{9n+4} − t_{3n+1}

**Proof :** We want to prove that the square of any odd multiple of 3 is the difference of two triangular numbers.

Since,

\(t_k=\frac{k(k+1)}{2}\), then

\(t_{9 n+4}=\frac{(9 n+4)(9 n+5)}{2}, t_{3 n+1}=\frac{(3 n+1)(3 n+2)}{2}\)Therefore,

\(t_{9 n+4}-t_{3 n+1}=\frac{\left(81 n^2+81 n+20\right)-\left(9 n^2+9 n+2\right)}{2}\)= \(\frac{72 n^2+72 n+18}{2}\) = 36n^{2} + 36n + 9

= 9(4n^{2} + 4n + 1) = 9(2n+1)^{2}

Thus, we proved that the square of any odd multiple of 3 is the difference of two triangular numbers specifically, that 9(2n+1)^{{2}} = t_{9n+4} − t_{3n+1}

**Page 16 Problem 5 Answer**

Here, we have to find two triangular numbers whose sum and difference are also triangular numbers.

From the list of the first ten triangular numbers, 1,3,6,10,15,21,28,36,45,55, if we look carefully, it is not hard to see that the number we searching for are A = 15 and B = 21.Really, A + B = 36 triangular number, B − A = 6 triangular number.

So, make a list of first ten triangular numbers i.e 0,1,3,6,10,15,21,28,36,45,55

Here, we have to find three successive triangular numbers whose product is a perfect square.

Product of three successive triangular numbers is given by,

\(\frac{n(n+1)}{2} \cdot \frac{(n+1)(n+2)}{2} \cdot \frac{(n+2)(n+3)}{2}\) = \(\frac{n(n+1)^2(n+2)^2(n+3)}{8}\)

for some positive integer n. We can conclude that 2 divide n+1 or n+2,then 4 must divide (n+1)^{2} or (n+2)^{2}.So that, if we find n such that n(n+3) divided by 2 gives the perfect square, our problem was solved. It is not hard to conclude that we need to chose n = 3

Really, for n = 3

\(\frac{3 \cdot 4 \cdot 4 \cdot 5 \cdot 5 \cdot 6}{8}\) = 3^{2} . 5^{2} . 4 = 30^{2} = 900

Thus, use that the product of the three successive triangular numbers is given by

\(\frac{n(n+1)}{2} \cdot \frac{(n+1)(n+2)}{2} \cdot \frac{(n+2)(n+3)}{2}\) = \(\frac{n(n+1)^2(n+2)^2(n+3)}{8}\) , for some positive integer n

Here, we have to find the three successive triangular numbers whose sum is a perfect square.

From the list of first ten triangular numbers 1,3,6,10,15,21,28,36,45,55, if we look carefully, it is not hard to see that the number we searching for are A = 15,B = 21 and C = 28 Really, A + B + C = 64 perfect square

Thus, make a list of first ten triangular numbers i.e 0,1,3,6,10,15,21,28,36,45,55

**Page 16 Problem 6 Answer**

Here, we have to prove that t_{4n(n+1)} is also a square if the triangular number t_{n} is a perfect square.

**Proof:** Assume

t_{n} = k^{2} = \(\frac{n(n+1)}{2}\), then 2k2 = n(n+1)

t_{4n(n+1)}= \(\frac{4 n(n+1)(4 n(n+1)+1)}{2}\)

= \(4 \cdot 2 k^2\left[4 n^2+4 n+1\right]\)

= 4k^{2}(2n+1)^{2}

= [2k(2n+1)]^{2}

Thus, we prove that t_{4n(n+1)} is also a square.

Here, we have to use part (a) to find three examples of squares that are also triangular numbers.

Using part (a) t_{1} = 1 is a perfect square. t4⋅1(1+1) = t_{8} = 36 is a perfect square.

\(t_{4 \cdot 8(8+1)}=t_{288}=\frac{288(288+1)}{2}\) = 41616 = 204^{2} is a perfect square.

Thus, we used part (a) to find three examples of squares that are also triangular numbers.

**Page 16 Problem 7 Answer**

Here, we have to show that the difference between the squares of two consecutive triangular numbers is always a cube.

**Proof :** We want to prove the difference between the squares of two consecutive triangular numbers is always a cube.

Since, \(t_{n+1}=\frac{(n+1)(n+2)}{2}, t_n=\frac{n(n+1)}{2}\)

∴ \(t_{n+1}^2-t_n^2=\frac{(n+1)^2(n+2)^2-(n+1)^2 n^2}{4}\)

= \(\frac{(n+1)^2\left[n^2+4 n+4=n^2\right]}{4}\) for n ≥ 1

= \(\frac{(n+1)^2(4 n+4)}{4}=(n+1)^3\)

Thus, show that the difference between the squares of two consecutive triangular numbers is always a cube.

**Page 16 Problem 8 Answer**

Here, we have to prove that the sum of the reciprocals of the first n triangular numbers is less than 2 that is,\(\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\cdots+\frac{1}{t_n}<2\). It’s given that \(\frac{2}{n(n+1)}=2\left(\frac{1}{n}-\frac{1}{n+1}\right)\)

**Proof:**

Note that,

\(\frac{1}{t_k}=\frac{1}{\frac{k(k+1)}{2}}=\frac{2}{k(k+1)}=2\left(\frac{1}{k}-\frac{1}{k+1}\right)\)∴ \(\frac{1}{1}+\frac{1}{3}+\cdots+\frac{1}{t_n}\)

= \(2\left(\frac{1}{1}-\frac{1}{2}\right)+2\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+2\left(\frac{1}{n}-\frac{1}{n+1}\right)\)

=\(2\left(\frac{1}{1}-\frac{1}{n+1}\right)=2\left(1-\frac{1}{n+1}\right)\)

Since,

\(n>0 \Rightarrow n+1>0 \Longrightarrow \frac{1}{n+1}>0 \Longrightarrow-\frac{1}{n+1}<0\) \(\Longrightarrow 1-\frac{1}{n+1}<1 \Longrightarrow 2\left(1-\frac{1}{n+1}\right)<2\)that is \(\frac{1}{1}+\frac{1}{3}+\cdots+\frac{1}{t_n}<2\)

Thus, we proved that sum of the reciprocals of the first n triangular numbers is less than 2 that is 1

\(\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\cdots+\frac{1}{t_n}<2\)

**Page 16 Problem 9 Answer**

Here, we have to establish the identity t_{{x}} = t_{{y}} + t_{{z}}, where \(x=\frac{n(n+3)}{2}+1 \quad y=n+1 \quad z=\frac{n(n+3)}{2}\) and n ≥ 1, thereby proving that there are infinitely many triangular numbers that are the sum of two other such numbers.

**Proof:**

R.H.S

= \(t_y+t_z=\frac{(n+1)(n+2)}{2}+\frac{\frac{n(n+3)}{2}\left[\frac{n(n+3)}{2}+1\right]}{2}\)

= \(\frac{2\left[\frac{(n+1)(n+2)}{2}\right]+\frac{n(n+3)}{2}\left[\frac{n(n+3)}{2}+1\right]}{2}\)

= \(\frac{2\left[\frac{n^2+3 n+2}{2}\right]+\frac{n(n+3)}{2}\left[\frac{n(n+3)}{2}+1\right]}{2}\)

= \(\frac{2\left[\frac{n(n+3)}{2}+1\right]+\frac{n(n+3)}{2}\left[\frac{n(n+3)}{2}+1\right]}{2}\)

= \(\frac{\left[\frac{n(n+3)}{2}+1\right]\left[\frac{n(n+3)}{2}+1+1\right]}{2}\)

= t_{x} = L.H.S

Thus, we prove that t_{{x}} = t_{{y}} + t_{{z}} where \(x=\frac{n(n+3)}{2}+1 \quad y=n+1 \quad z=\frac{n(n+3)}{2}\)

Here, we have to find three examples of triangular numbers that are sums of two other triangular numbers.

Using part (a):

n = 1 : t_{3} = t_{2} + t_{2}, or 6 = 3 + 3

n = 2 : t_{6} = t_{3} + t_{5}, or 21 = 6 + 15

n = 3 : t_{10} = t_{4} + t_{9}, or 55 = 10 + 45

Thus, we find three examples of triangular numbers that are sums of two other triangular numbers, they are

n = 1 : t_{3} = t_{2} + t_{2}, or 6 = 3 + 3

n = 2 : t_{6} = t_{3} + t_{5}, or 21 = 6 + 15

n = 3 : t_{10} = t_{4 }+ t_{9}, or 55 = 10 + 45

**Page 16 Problem 10 Answer**

Here, we have to prove that p_{n} = \(\frac{n(3 n-1)}{2}\), n ≥ 1. It’s given that each of the numbers 1, 5 = 1 + 4, 12 = 1 + 4 + 7, 22 = 1 + 4 + 7 + 10,… represents the number of dots that can be arranged evenly in a pentagon:

Let p_{n} denotes nth pentagonal number where, p_{1} = 1

p_{n} = p_{n-1} + 3n – 2, n ≥ 2

We need to prove that n th pentagonal number is given by formula p_{n} = \(\frac{n(3 n-1)}{2}\) equation (1) for each positive integer n for which Eq (1) holds.

For n = 1 eq (1) is true, so that 1 belongs to the set S.

We assume that Eq (1) is true for a fixed integer k, so that for this k, p_{k} = \(\frac{k(3 k-1)}{2}\) and we attempt to prove that validity of the formula for k + 1. Using the definition of the pentagonal numbers we obtain that, p_{k+1} = p_{k} + 3(k+1)-2

Next, from hypothesis that Eq (1) is correct for k we have that

p_{k+1} = p_{k} + 3(k+1)-2

= \(\frac{k(3 k-1)}{2}+\frac{6(k+1)-4}{2}\)

= \(\frac{3 k^2-k+6 k+6-4}{2}\)

= \(\frac{3 k^2+5 k+2}{2}\)

= \(\frac{3 k^2+3 k+2 k+2}{2}\)

= \(\frac{3 k(k+1)+2(k+1)}{2}\)

= \(\frac{(k+1)(3(k+1)-1)}{2}\)

But this says that Eq (1) holds when n = k + 1,putting the integer k+1 in S so, that k+1 is in S whenever k is in S.According to the induction principle, S must be the set of all positive integers.

Thus, we use the definition of the pentagonal numbers and the induction principle to prove p_{n} = \(\frac{n(3 n-1)}{2}\), n ≥ 1

**Page 16 Problem 11 Answer**

Here, we have to verify the relations between the pentagonal, square, and triangular numbers: p_{n} = t_{n-1 }+ n^{2}

Let n be fixed arbitrary integer such that n ≥ 2, then

\(t_{n-1}+n^2 \stackrel{(1)}{=} \frac{n(n+1)}{2}+n^2\)= \(\frac{n^2+n+2 n^2}{2}\)

= \(\frac{n(3 n+1)}{2}\)

= p_{n}

So, that this relation is correct.

Thus, we verified the relations between the pentagonal, square, and triangular numbers ; p_{n} = t_{n-1} + n^{2}

Here, we have to verify the relations between the pentagonal, square, and triangular numbers: pn = 3t_{n-1} + n = 2t_{n-1} + t_{n}

Let n be fixed arbitrary integer such that n ≥ 2,then,

\(3 t_{n-1}+n \stackrel{(1)}{=} \frac{n(n-1)}{2}+n\)= \(\frac{3 n^2-3 n+2 n}{2}\)

= \(\frac{3 n^2-n}{2}\)

= \(\frac{n(3 n+1)}{2}\)

= \(p_n\)

= \(\frac{n(3 n+1)}{2}\)

= \(\frac{3 n^2-n}{2}\)

= \(\frac{2 n^2-2 n+n^2+n}{2}\)

= \(2 \frac{n(n-1)}{2}+\frac{n(n+1)}{2}\)

= 2t_{n-1} + t_{n}

So, the these relations are correct

Thus, we verified the relations between the pentagonal, square, and triangular numbers: p_{n} = 3t_{n-1} + n = 2t_{n-1} + t_{n}