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Number Theory Level 5

⌊ a 2 b 2 ⌋ + ⌊ b 2 a 2 ⌋ = ⌊ a 2 + b 2 a b ⌋ + a b \left \lfloor \frac { { a }^{ 2 } }{ { b }^{ 2 } } \right \rfloor +\left \lfloor \frac { { b }^{ 2 } }{ { a }^{ 2 } } \right \rfloor = \left \lfloor \frac { { a }^{ 2 }+{ b }^{ 2 } }{ ab } \right \rfloor + ab

If there exists a least pair of natural numbers ( x , y ) (x, y) to the equation above for positive natural a a and b b , find x + y x + y .

Notation : ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 3.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

Kushal Bose
Sep 18, 2016

Let a = b k + r a=bk+r where k ∈ N k \in N and 0 ≤ r ≤ b 0 \leq r \le b

Consider r = 0 r=0 and putting this in the given equation:

k 2 + 0 = k + b 2 k k^2 + 0= k+b^2 k as b a < 1 \frac{b}{a} \lt 1

k = 1 + b 2 k=1+b^2 as k ≠ 0 k \neq 0

Now a = b ( 1 + b 2 ) a=b(1+b^2)

This relation will give many solutions.For minimum solution putting b = 1 b=1 we get a = 2 a=2

Minimum ordered pair is ( 2 , 1 ) (2,1)

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