Minimum value

Algebra Level 3

If m m and n n are consecutive positive integers such that n 2 m 2 > 20 n^2-m^2 >20 then what is the minimum value of m 2 + n 2 m^2+n^2 ?

221 189 179 220 199

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1 solution

Zee Ell
Mar 10, 2017

If n and m are consecutive positive integers, then m = n - 1 (if we take the inequality into account, then we can see that n > m).

n 2 m 2 > 20 n^2 - m^2 > 20

n 2 ( n 1 ) 2 > 20 n^2 - (n - 1)^2 > 20

2 n 1 > 20 2n - 1> 20

n > 10.5 n > 10.5

n 11 n ≥ 11

m 10 m ≥ 10

Hence, the minimum value is:

1 1 2 + 1 0 2 = 221 11^2 + 10^2 = \boxed {221}

Nice solution. Thank you.

Hana Wehbi - 4 years, 3 months ago

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