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 xx and yy such that x2+84x+2008=y2x^2 + 84x + 2008 = y^2. Find x+yx + y.
Solution
Answer: 80
Step 1: Complete the square. Since 84/2=4284 / 2 = 42,
x2+84x+2008=(x2+84x+1764)+244=(x+42)2+244=y2.x^2 + 84x + 2008 = (x^2 + 84x + 1764) + 244 = (x + 42)^2 + 244 = y^2.
 Step 2: Difference of squares.
y2−(x+42)2=244  ⟹  [y−(x+42)][y+(x+42)]=244.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 2y2y, which is even, so both factors have the same parity. As 244=22⋅61244 = 2^2 \cdot 61, if both were odd their product would be odd; hence both are even. The only factorisation of 244244 into two positive even factors is 244=2×122244 = 2 \times 122. Step 4: With y+(x+42)>y−(x+42)y + (x + 42) > y - (x + 42),
y−(x+42)=2,y+(x+42)=122.y - (x + 42) = 2, \qquad y + (x + 42) = 122.
 Adding gives 2y=1242y = 124, so y=62y = 62; subtracting gives 2(x+42)=1202(x + 42) = 120, so x=18x = 18. Step 5: Therefore x+y=18+62=80x + y = 18 + 62 = 80. Answer: x+y=80x + y = 80.
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Algebra (Olympiad) · medium
The all pairs of real numbers (x,y)(x, y) satisfying (2x+1)2+y2+(y−2x)2=13(2x + 1)^2 + y^2 + (y - 2x)^2 = \dfrac{1}{3}.
Algebra (Olympiad) · medium
The number of all real pairs (a,b)(a, b) satisfying a2+b2=25a^2 + b^2 = 25 and 3(a+b)−ab=153(a + b) - ab = 15.
Algebra (Olympiad) · hard
Let {an}\{a_n\} be a sequence of positive integers such that a1=a2=2a_1=a_2=2. For n≥1n\ge1, an+2an−an+12=an+1an.a_{n+2}a_n-a_{n+1}^2=a_{n+1}a_n. Determine the value of a1a2a1032 a7\dfrac{a_1a_2a_{10}}{32\,a_7}.
Algebra (Olympiad) · hard
Consider the polynomial (x2+1)(x2+4)(x2−2x+2)(x2+2x+2).(x^2+1)(x^2+4)(x^2-2x+2)(x^2+2x+2). Let f(x)=(x4+2x2+2x)2+(P(x))2f(x)=\big(x^4+2x^2+2x\big)^2+\big(P(x)\big)^2, where P(x)P(x) is a cubic polynomial. Find the value of P(1) P(2)P(1)\,P(2)

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath