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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


But try the problem first…

Answer: is 80.

Source
Suggested Reading

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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