MHB Max Value of a: Positive Integer Solutions

[anemone]

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

 

[Greg]

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

Spoiler

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 $68$.

 

[anemone]

Spoiler

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 $68$.

Spoiler

Nice try greg1313, but sorry, your answer isn't correct..:(

 

[Greg]

Spoiler

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!

 

[kaliprasad]

Spoiler

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 = $2500$

 

[anemone]

Spoiler

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)