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Equations and Integers | AIME I, 2008 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Equations and Integers.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Equations and integers.

Equations and integers – AIME I, 2008


There exists unique positive integers x and y that satisfy the equation \(x^{2}+84x+2008=y^{2}\)

  • is 107
  • is 80
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Integers

Check the Answer


Answer: is 80.

AIME I, 2008, Question 4

Elementary Number Theory by David Burton

Try with Hints


\(y^{2}=x^{2}+84x+2008=(x+42)^{2}+244\) then 244=\(y^{2}-(x+42)^{2}=(y-x-42)(y+x+42)\)

here 244 is even and 244=\(2^{2}(61)\)=\( 2 \times 122\) for \(x,y \gt 0\)

(y-x-42)=2 and (y+x+42)=122 then y+x=80 and x=18 y=62.

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