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April 10, 2020

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.

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


First hint

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

Second Hint

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

Final Step

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

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