Question

prove that there exist no natural number m and n such that m^2=n^+2002

Answer 1

Solution :

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Given :

To prove there exist no natural number m & n such that

⇒ m² = n² + 2002

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Proof :

We can right the given format in the following ways,.

⇒ m² = n² + 2002

⇒ m² - n² = 2002

Hence, the difference of 2 squares of natural numbers is 2002,.

We know that,

⇒ a² - b² = (a + b)(a - b)

Hence,

The given equation can also be written as,

⇒ (m + n)(m - n) = 2002

For a product of two integers to be even,.

The atleast 1 integer should also be even (2 x 3 = 6),etc,.

⇒ Either m + n is even (or) m - n is even or both are even numbers

If m + n is even Then,

m = even number - n

 If m is odd, then, n is also odd ( Odd + Odd = even)

If m is even then,

n must be even,.

∴ m & n can be either even or odd,. but, both are same kind of number,.

There is no such number to have a difference = 2002

By prime factorization,

⇒ 2002

⇒ 2 x 7 x 11 x 13 ..(it must have 2 2s as factor to have such possible difference ),.

Hence,

There is no value for m + n & m - n to give product as 2002,.

Hence, There is no possible natural numbers of m & n to be in the given form,/


                                  Hence Proved,.

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Answer 2

Answer:

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