Question

If two positive integers m and n , both bigger than 1 satisfy the equation m²+2005²=n²+2004²
Then find the value of m+n-200

Answer 1

Answer:

11  

Step-by-step explanation:

m²+2005²=n²+2004², m,n > 1

2005² - 2004² = n² -  m²  

(2004 + 1) ² - 2004² = (n + m) (n - m)

(2004² + 2(2004 x 1) + 1²) - 2004² = (n + m) (n - m)

2(2004) + 1 + 2004² - 2004² = (n + m) (n - m)

4008 + 1 = (n + m) (n – m)

4009 x 1 = (n + m) (n - m)

Now, 4009 has 2 factors, 19, 211

211 x 19 = (n + m) (n - m)

211 is a prime so it can’t be factored farther

m + n = 211

211 – 200 = 11.