Max Value of a: Positive Integer Solutions

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Discussion Overview

The discussion revolves around finding the maximum value of a positive integer \( a \) such that both \( a \) and \( \sqrt{a^2 + 204a} \) are positive integers. The scope includes mathematical reasoning and problem-solving related to integer solutions.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Post 1 presents the problem of determining the maximum value of \( a \) under the given conditions.
  • Post 2 reiterates the same problem statement, indicating a potential emphasis on the challenge.
  • Post 3 expresses a correction or disagreement with a previous claim, though it does not specify the nature of the error.
  • Post 4 offers praise to another participant, suggesting a positive interaction but does not contribute to the mathematical discussion.

Areas of Agreement / Disagreement

The discussion does not reach a consensus, as there is at least one disagreement indicated by Post 3, and the nature of the claims remains unresolved.

Contextual Notes

There are no explicit assumptions or definitions provided, and the mathematical steps leading to a solution are not detailed.

Who May Find This Useful

Participants interested in integer solutions, mathematical problem-solving, and those exploring conditions for positive integers may find this discussion relevant.

anemone
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If both $a$ and $\sqrt{a^2+204a}$ are positive integers, find the maximum value of $a$.
 
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anemone said:
If both $a$ and $\sqrt{a^2+204a}$ are positive integers, find the maximum value of $a$.

As $\sqrt{a^2+204a}\equiv a\sqrt{1+\dfrac{204}{a}},\quad1+\dfrac{204}{a}$ must also be a perfect square. As $204$ has factors $1,2,3,4,6,12,17,34,51,68,102,204$ and $1+\dfrac{204}{68}=4$ whereas $a=102$ and $a=204$ do not give perfect squares, the maximum value of $a$ is $68$.
 
greg1313 said:
As $\sqrt{a^2+204a}\equiv a\sqrt{1+\dfrac{204}{a}},\quad1+\dfrac{204}{a}$ must also be a perfect square. As $204$ has factors $1,2,3,4,6,12,17,34,51,68,102,204$ and $1+\dfrac{204}{68}=4$ whereas $a=102$ and $a=204$ do not give perfect squares, the maximum value of $a$ is $68$.

Nice try greg1313, but sorry, your answer isn't correct..:(
 
greg1313 said:
As $\sqrt{a^2+204a}\equiv a\sqrt{1+\dfrac{204}{a}},\quad1+\dfrac{204}{a}$ must also be a perfect square. As $204$ has factors $1,2,3,4,6,12,17,34,51,68,102,204$ and $1+\dfrac{204}{68}=4$ whereas $a=102$ and $a=204$ do not give perfect squares, the maximum value of $a$ is $68$.
Incorrect! Sorry about that! :o
 
let $\sqrt{a^2+204a} = y$
So $y^2 = a^2 + 204a$
or $y^2 = a^2 + 204 a + 102^2 - 102^2 = (a+102)^2- 102^2$
or $(a+102)^2-y^2 = 102^2$
or $(a+102+y)(a+102-y) = 102^2$
now $(a+102+y)$ and $(a+102-y)$ both should be even (as product is even) and for a to be maximum $a+102+y$
should be maximum and $(a+102-y)$ should be minumum so say $(a+102-y) =2$ and we get
$(a+102+y) = 102 * 51$ $(a+102-y) = 2$
adding $2a + 204 = 102 * 51 + 2$ or a = $2500$
 
kaliprasad said:
let $\sqrt{a^2+204a} = y$
So $y^2 = a^2 + 204a$
or $y^2 = a^2 + 204 a + 102^2 - 102^2 = (a+102)^2- 102^2$
or $(a+102)^2-y^2 = 102^2$
or $(a+102+y)(a+102-y) = 102^2$
now $(a+102+y)$ and $(a+102-y)$ both should be even (as product is even) and for a to be maximum $a+102+y$
should be maximum and $(a+102-y)$ should be minumum so say $(a+102-y) =2$ and we get
$(a+102+y) = 102 * 51$ $(a+102-y) = 2$
adding $2a + 204 = 102 * 51 + 2$ or a = $2500$

Bravo, kaliprasad!(Cool)
 

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