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Home / Elementary Number Theory / Show that the cube of any integer is of the form 7k or 7k ± 1.

Show that the cube of any integer is of the form 7k or 7k ± 1.

Proof: Let A be an integer. We have to show that the cube of A is 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

So from above we can say that the cube of any integer is of the form 7k or 7k ± 1. (Showed)

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