Algebra (Olympiad)hardFree

Solve x²+84x+2008=y², Find x+y | IOQM

JEE Maths question with a full step-by-step solution.

Question

There exist unique positive integers

x

and

y

such that

x^2 + 84x + 2008 = y^2

. Find

x + y

Solution

Answer: 80

Step 1: Complete the square. Since

84 / 2 = 42

x^2 + 84x + 2008 = (x^2 + 84x + 1764) + 244 = (x + 42)^2 + 244 = y^2.

Step 2: Difference of squares.

y^2 - (x + 42)^2 = 244 \implies \big[y - (x + 42)\big]\big[y + (x + 42)\big] = 244.

Step 3: Parity. The sum of the two factors is

2y

, which is even, so both factors have the same parity. As

244 = 2^2 \cdot 61

, if both were odd their product would be odd; hence both are even. The only factorisation of

244

into two positive even factors is

244 = 2 \times 122

. Step 4: With

y + (x + 42) > y - (x + 42)

y - (x + 42) = 2, \qquad y + (x + 42) = 122.

Adding gives

2y = 124

, so

y = 62

; subtracting gives

2(x + 42) = 120

, so

x = 18

. Step 5: Therefore

x + y = 18 + 62 = 80

. Answer:

x + y = 80

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