Let k N Prove that a positive integer n is a kth power if an

Let k N. Prove that a positive integer n is a k-th power if and only if every exponent appearing in the prime-power factorization of n is a multiple of k. Then use this to find a positive integer n for which n/3 is a square and n/7 is a cube.

Solution

Lets Understand the question first:-

Let a number n which is a k-th power.

We have to take perfect powered \'n\'. (E.g 36 is 6² so n is 36 || 8 is 2³ so n is 8)

Lets Evaluate 2 Examples:-

1. 216 = 6³ = (2*3)³ = 2³ * 3³ (n is kth power, so n=216, k=3, )

so Prime Factorization of 216 = 2³ * 3³ ,

(Exponent of 2)/k = 3/k = 3/3 = 1 . Exponent of 2 i.e,3 is a multiple of k

(Exponent of 3)/k = 3/k = 3/3 = 1   Exponent of 3 i.e,3 is a multiple of k

2. 144 = 12² = (4*3)²= (2²* 3)² = 2^2*2 * 3² = 2⁴ * 3² (n is kth power, so n=144, k=2, )

so Prime Factorization of 144 = 2⁴* 3²,

(Exponent of 2)/k = 4/k = 4/2 = 2 . Exponent of 2 i.e,4 is a multiple of k

(Exponent of 3)/k = 2/k = 2/2 = 1   Exponent of 3 i.e,2 is a multiple of k

Proof:-

So, n   = (n^k)^1/k , Let x₁, x₂ ,x₃ ......., x_n Prime Factors of n.

Now, n = [{(x₁)^a * (x₂)^b * (x₃)^c....}^k]^1/k,a,b,c... are powers or exponent of prime factors of n generated after factorization

n = [{(x₁)^a*k * (x₂)^b*k * (x₃)^c*k....}]^1/k , So we can see that Every Prime factor exponent of n is a multiple of k.

Part [B] :- To Find A positive Integer n for which n/3 is a square and n/7 is a cube.

This is slightly Tricky:-

As n/3 is a square and n/7 is a cube,

Since n is an INTEGER, So we cannot have fractions here. For making n/3 a square or n/7 a cube, number should already have 3 , 7 as denominators which is not possible as n is Integer.

Number n is definitely a multiple of 3 and 7. As we need n/3 and n/7.

Lets take n = 3^a * 7^b a,b are exponents of 3 and 7

Case 1: As we need 3^a/3 as perfect square so a could be, a = 1,3,5,7,9,11...... as (a-1) should be divisible by 2.

Case 2: As we need 7^b/7 as perfect cube so b could be, b = 1,4,7,10,13....... as (b-1) should be divisible by 3.

Case 3: n/3 = Perfect Square Number = x,, x = n/3 = 3^a-1 * 7^b.

For x to be perfect square we need the even exponent of 7. b could be 2,4,6,8..

Case 4 : n/7 = Perfect Cube NUmber = y ,,, y = n/7 = 3^a * 7^b-1

For y to be perfect cube we need 3\'s exponent in n as multiple of 3. a could be 3,6,9,12

Combining Case 1 and Case 4

a = 1,3,5,7,9,11 & a = 3,6,9,12, 15, 21.....

Net Result is, a = 3 , 9 , 15, 21, 27... in arithmetic progression of +6

Combining Case 2 and Case 3

b = 1,4,7,10,13, 16, 19, 22...... & b = 2,4,6,8,10,12,14,16,18, 20, 22...

Net Result is b = 4, 10, 16, 22... in an Arithmetic Progression of +6

Possible Values of n are:-

3^a * 7^b where, a = 3,9,15,21,27.... and b = 4,10,16,22.......

so a = 3r and b = (3r+1) where r is positive odd numbers

n = 3^3r * 7^(3r+1) , r = Positive odd numbers

For All Values of n :- we will have to consider other possible factors of n. i.e n = 3^a*7^b*N^c

Factor Say N should have the exponent at least 6 as it has to be a perfect square as well as perfect cube for that exponent = 2*3 = 6. c = 6*whole numbers. Whole Numbers = 0,1,2,3,4....

So All Values of n = N^6*v * 3^3r* 7^3r+1

v = All Whole Numbers 0,1,2,3,4,5,6,7.....

r = Positive odd numbers 1,3,5,7,....