Theory of Numbers - Problem Set Basic

1) If a² - b²  is a prime number, show that a² - b²  = a +b, where a and b are natural numbers

a² - b²  = (a-b) (a+b)

∵ (a² - b² ) is a prime number
∴ one of the two factors = 1
∵ a -b = 1 [∵ a -b < a + b]

∵ the only divisor of a prime number are 1 and itself

a² - b²  = 1 (a +b)

or a² - b²  =  a + b

example 3² - 2² = 5 (which is  a prime)

3² - 2² = 3 + 2,    3,2 𝝐 N

2) Prove  that sqaure of each odd number is of the form 8j+1 

Let n = 2m + 1 be an odd number

n² = (2m+1)² = 4m² + 4m + 1
= 4m (m+1) + 1

Now m (m+1) being product of two consecutive integers, is divisible by 2! = 2

m(m+1) = 2j

⇒ n² = 4 (2j) +1 = 8j+1

3) Show that sum of an interger and its square is even 
Let n be any integer
so, we have to prove that n²  + n is even
⇒  n²  + n = n (n+1) which is product of two consecutive number n and n+1 and hence divisible by 2!  = 2

Hence,  n²  + n is an even number

4) If n is an integer. Prove that product n( n²  - 1 ) is a multiple of 6

n( n²  - 1 ) = n(n-1)(n+1) = (n-1) n (n+1)
which being the product of three consecutive integers is divisible by 3! = 6

∴ n (n²  - 1 ) is divisible by 6
i.e.  n (n²  - 1 ) is a multiple of 6

5. Prove 8ⁿ - 3ⁿ is divisible by 5

8ⁿ - 3ⁿ = ( 8 - 3) (8ⁿ⁻¹ + 8ⁿ⁻² . 3 +... 3ⁿ⁻¹)

or 8ⁿ - 3ⁿ = ( 5) (8ⁿ⁻¹ + 8ⁿ⁻² . 3 +... 3ⁿ⁻¹)

∴ 8ⁿ - 3ⁿ is divisible by 5

6. Prove that n⁵ - n is divisible by 5

n⁵ - n = n(n-1)(n+1)(n²  + 1)

If the integer n  terminates with one of the digits 0, 1, 4, 5, 6, or 9 then one of the first three
is divisible by 5

If n ends in one of the digits 2, 3, 7 or 8 then n²  ends in 4 or 9, and in this event n²  + 1 is divisible by 5

7. Prove that n⁷ - n is divisible by 7

8. Prove that n¹¹ - n is divisible by 11

9. Prove that n¹³ - n is divisible by 13

10. Prove  

So, 25, 195 is divisible by . The difference shown in the parentheses is divisible by

27,195 - 10,152 = 735 = 3.5.7²

. Hence, N is divisible by 5.7²