BIGtheme.net http://bigtheme.net/ecommerce/opencart OpenCart Templates
Friday , July 28 2017
Home / Elementary Number Theory / Verify that if an integer is simultaneously a square and a cube, then it must be either of the form 7k or 7k + 1

# Verify that if an integer is simultaneously a square and a cube, then it must be either of the form 7k or 7k + 1

Proof: Let A be an integer. We have to show that the cube and square of A is of the form 7k or 7k + 1.

First we show that, a square is either of the form 7k or 7k + 1.

According to the division algorithm, we have

A = 7q + r, 0 ≤ 0 < 6

For, r = 0, we have

A = 7q

⟹ A2 = (7q)2 = 49q2 = 7. 7q2

Let, k = 7q2

∴ A2 = 7k

For, r = 1, we have

A = 7q + 1

⟹ A2 = (7q + 1)2

= 49q2 + 14q + 1

= 7(7q2 + 2q) + 1

Let, k = 7q2 + 2q

∴ A2 = 7k + 1

For, r = 2, we have

A = 7q + 2

⟹ A2 = (7q + 2)2

= 49q2 + 28q + 4

= 7(7q2 + 4q) + 4

Let, k = 7q2 + 4q

∴ A2 = 7k + 4

For, r = 3, we have

A = 7q + 3

⟹ A2 = (7q + 3)2

= 49q2 + 42q + 9

= 49q2 + 42q + 7 + 2

= 7(7q2 + 6q + 1) + 2

Let, k = 7q2 + 6q + 1

∴ A2 = 7k + 2

For, r = 4, we have

A = 7q + 4

⟹ A2 = (7q + 4)2

= 49q2 + 56q + 16

= 49q2 + 56q + 14 + 2

= 7(7q2 + 8q + 2) + 2

Let, k = 7q2 + 8q + 2

∴ A2 = 7k + 2

For, r = 5, we have

A = 7q + 5

⟹ A2 = (7q + 5)2

= 49q2 + 70q + 25

= 49q2 + 70q + 21 + 4

= 7(7q2 + 10q + 3) + 4

Let, k = 7q2 + 10q + 3

∴ A2 = 7k + 4

For, r = 6, we have

A = 7q + 6

⟹ A2 = (7q + 6)2

= 49q2 + 84q + 36

= 49q2 + 84q + 35 + 1

= 7(7q2 + 12q + 5) + 1

Let, k = 7q2 + 12q + 5

∴ A2 = 7k + 1

We have, A2 = 7k, if r = 0

A2 = 7k + 1, if r = 1, 6

A2 = 7k + 2, if r = 3, 4

And A2 = 7k + 4, if r = 2, 5

Now we show that, a cube is either of the form 7k or 7k + 1.

According to the division algorithm we can write,

A = 7q + r, 0 ≤ r < 7

When r = 0, then

A = 7q

⟹ A3 = (7q)3 = 73q3 = 7(72q3)

Let, k = 72q3

∴ A3 = 7k

When r = 1, then

A = 7q + 1

⟹A3 = (7q + 1)3

= 73q3 + 3. 72 q2. 1 + 3. 7q. 1 + 1

= 7(72q3 + 3. 7q2 + 3q) + 1

Let, (72q3 + 3. 7q2 + 3q) = k

∴ A3 = 7k + 1

When r = 2, then

A = 7q + 2

⟹A3 = (7q + 2)3

= 73q3 + 3. 72 q2. 2 + 3. 7q. 22 + 23

= 73q3 + 3. 72 q2. 2 + 3. 7q. 4 +8

= 73q3 + 3. 72 q2 .2+ 3. 7q.4 + 7 + 1

= 7(72q3 + 3. 7q2 .2+ 3q.4 + 1) + 1

Let, (72q3 + 3. 7q2 + 3q.4 +1) = k

∴ A3 = 7k + 1

When r = 3, then

A = 7q + 3

⟹A3 = (7q + 3)3

= 73q3 + 3. 72 q2. 3 + 3. 7q. 32 + 33

= 73q3 + 3. 72 q2. 3 + 3. 7q. 9 +27

= 73q3 + 3. 72 q2 .3+ 3. 7q.9 + 28 – 1

= 7(72q3 + 3. 7q2 .3+ 3q.9 +4) – 1

Let (72q3 + 3. 7q2 .3+ 3q.9 +4) = k

∴ A3 = 7k – 1

When r = 4, then

A = 7q + 4

⟹A3 = (7q + 4)3

= 73q3 + 3. 72 q2. 4 + 3. 7q. 42 + 43

= 73q3 + 3. 72 q2. 4 + 3. 7q. 16 +64

= 73q3 + 3. 72 q2 .4+ 3. 7q. 16 + 63 +1

= 7(72q3 + 3. 7q2 .4+ 3q.16+9) – 1

Let, (72q3 + 3. 7q2 .4+ 3q.16+9) = k

∴ A3 = 7k – 1

When r = 5, then

A = 7q + 5

⟹A3 = (7q + 5)3

= 73q3 + 3. 72 q2. 5 + 3. 7q. 52 + 53

= 73q3 + 3. 72 q2. 5 + 3. 7q. 25 +125

= 73q3 + 3. 72 q2 .5+ 3. 7q. 25 + 126 -1

= 7(72q3 + 3. 7q2 .5+ 3q.25+18) – 1

Let, (72q3 + 3. 7q2 .5+ 3q.25+18) = k

∴ A3 = 7k – 1

When r = 6, then

A = 7q + 6

⟹A3 = (7q + 6)3

= 73q3 + 3. 72 q2. 6 + 3. 7q. 62 + 63

= 73q3 + 3. 72 q2. 6 + 3. 7q. 36 +216

= 73q3 + 3. 72 q2 .6+ 3. 7q. 36 + 217 – 1

= 7(72q3 + 3. 7q2 .6+ 3q.36+31) – 1

Let, (72q3 + 3. 7q2 .6+ 3q.36+31) = k

∴ A3 = 7k – 1

We have A3 = 7k if r = 0

A3 = k + 1 if r = 1, 2, 4

And A3 =7k + 6 if r = 3, 5, 6

Thus, A2 of form 7k, 7k + 1, 7k + 2, 7k + 4 and A3 is of form 7k, 7k + 1, 7k + 6

By uniqueness part of division algorithm A must be either of form 7k or 7k + 1. (Proved)

## Application of Euclidian’s algorithm in Diophantine equation

Problem-1: Which of the following Diophantine equations cannot be solved –   a) 6x + ...