(D)ragon(b)all (Z)

Number Theory Level 4

( 9 d + 80 b ) 2 − 80 ( 9 b + d ) 2 = 1 (9d+80b)^2-80(9b+d)^2=1

If d , b ∈ Z + ∪ { 0 } d,b\in \mathbb{Z}^{+} \cup \{0\} , then how many solutions does the above equation have?

2 Infinte Can't say 1 4

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

Akshat Sharda
Sep 23, 2017

( 9 d + 80 b ) 2 − 80 ( 9 b + d ) 2 = 1 ⇒ ( d ) 2 − 80 ( b ) 2 = 1 (9d+80b)^2-80(9b+d)^2=1 \Rightarrow (d)^2-80(b)^2=1

Note that the second eqation [ ( d ) 2 − 80 ( b ) 2 = 1 ] [(d)^2-80(b)^2=1] looks similar to the first equation [ ( 9 d + 80 b ) 2 − 80 ( 9 b + d ) 2 = 1 ] [(9d+80b)^2-80(9b+d)^2=1] .

Observe that ( d , b ) = ( 9 , 1 ) (d,b)=(9,1) satisfies the second equation. Therefore putting these values in first equation also satisfies, therefore,

( 9 × 9 + 80 ) 2 − 80 ( 9 + 9 ) 2 = 1 ⇒ ( 161 ) 2 − 80 ( 18 ) 2 = 1 (9×9+80)^2-80(9+9)^2=1 \Rightarrow (161)^2-80(18)^2=1

Now this is similar to second equation, therefore, ( d , b ) = ( 161 , 18 ) (d,b)=(161,18) also satisfies. Now we can put these in the first equation and compare it to second and keep going.

Therefore, infinite solutions.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You can just simplify the LHS to get d 2 − 80 b = 1 d^2- 80b = 1 , which resembles a Pell's equation .

What's left to do is to find a trivial a solution, 9 2 − 80 ⋅ 1 = 1 9^2 - 80\cdot 1 = 1 . Since we've found one solution, then (by the properties of Pell's equation), there are infinitely many solutions.

Pi Han Goh - 3 years, 8 months ago

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