Determine the set of integers n for which n2 + 19n +92 is a square.
Answers
Correct Question :
Determine the set of Integers n for which n² + 19n + 92 is a perfect square.
Solution :
We have to here determine the set of Integers n such that n² + 19n + 92 is a perfect square.
Suppose that this is equal to x²
> n² + 19n + 92 = x²
Multiplying by 4 on both sides to complete the square :
> 4n² + 76n + 368 = 4x²
> 4n² + 76n + 361 + 7 = 4x²
> 4n² + 76n + 361 = 4x² - 7
> ( 2n + 19)² = 4x² - 7
let 2n + 19 = c
> c² = 4x² - 7
> 4x² - c² = 7
> (2x + c)(2x - c) = 7
There are two possible cases :
• 2x + c = 7 & 2x - c = 1
• 2x + c = 1, 2x - c = 7
In both cases, adding up the equations we get
> 4x = 8
> x = 2 & c = ±3
Let's go back to
(2n + 19)² = 4x² - 7
Placing x = 2
> (2n + 19)² = 4 × 4 - 7
> (2n + 19)² = 9
From here we get n = -8 and -11.
These are the two sets of integral values of n satisfying the above condition.
This is the required answer.
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n² + 19n + 92 = m²
Multiplying throughout with
4 : 4n² + 76n + 368 = 4m²
=> (2n + 19)² + 7 = 4m²
=> 4m² – (2n – 19)² = 7
=> (2m + 2n + 19) . (2m – 2n – 19) = 7
Factors of 7 are: (1,7),(7,1),(–1,–7),(–7,–1)
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case - 1: => 2m + 2n + 19 = 7; 2m – 2n – 19 = 1
solving for m, n : m = 2 ; n = – 8
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case - 2: => 2m + 2n + 19 = 1; 2m – 2n – 19 = 7
solving for m, n : m = 2 ; n = – 11
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case - 3: => 2m + 2n + 19 = – 7; 2m – 2n – 19 = – 1
solving for m, n : m = – 2 ; n = – 11
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case - 4: => 2m + 2n + 19 = – 1; 2m – 2n – 19 = – 7
solving for m, n : m = – 2 ; n = – 8