There exist natural numbers, m & n such that `m^2 = n^2 + 2010.`

Prove that there exist no natural numbers, m and n such that m^(2)=n^(2)+2002.

Prove that there exist no natural numbers, m and n such that m^(2)=n^(2)+2002.

Consider the following statements : 1. There exists a positive real number m such that cos x = 2^(m +1) . 2. mn ge m+n for all m, n belonging to set of natural numbers. Which of the above statements is/are correct?

The A.M of first n natural numbers is

What is the largest value of m for which there exists a real value of n such that m^(2) = 196 -n^2 ?

The A . M first 'n' natural numbers is

Perfect square or square number A natural number n is called a perfect square or a square number if there exists a natural number m such that n=m^(2)

Can there exist natural numbers m and n for which 15^m 16^n can end with zero?

If m and n are the two natural number such that m^(n)=25 , then find the value of n^(m) :-

If m and n are the two natural number such that m^(n)=25 , then find the value of n^(m) :-