Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I came across this question during a Math Olympiad Competition. I need help with solving the question. Thanks.

share|improve this question
    
All numbers are either of the form : $4k, 4k+1, 4k+2,4k+3$ and the squares of the numbers are always $4k, 4k+1$. $n^2+204n$ is a perfect square. Hence, it must be of the form $4k$ or $4k+1$ Now, try to solve for $n^2+204n \equiv 1 \mod 4 $ and $n^2+204n \equiv 0 \mod 4 $ or $n^2+204n-1 \equiv 0 \mod 4 $ and $n(n+204) \equiv 0 \mod 4 $ –  VHP Jun 3 at 16:34

3 Answers 3

up vote 10 down vote accepted

Let $$m^2=n^2+204n$$ $$k^2=n^2+204n+10404=(n+102)^2$$ Then $$(k-m)(k+m)=2^2\cdot 3^2\cdot 17^2$$ Since $k-m$ and $k+m$ have the same parity, they must be even. Write:

$$\frac{k-m}2\frac{k+m}2=3^2\cdot17^2$$

Since $k>m$ there are only two options: $k-m=2$ and $k-m=18$. The former gives $m=2600$ and the latter $m=280$. Thus we are interested in the former, which gives $n=2500$.

share|improve this answer
1  
How do you get to $$(k-m)(k+m)=2^2\cdot 3^2\cdot 17^2$$ ? as in how do you get the $$2^2\cdot 3^2\cdot 17^2$$? –  snivysteel Jun 3 at 16:46
1  
Substract the two fisrt equations to get $k^2-m^2=10404$, now factor $10404$ and $k^2-m^2$. –  ajotatxe Jun 3 at 16:50
    
@ajotatxe How did you know to set k using: $$k^2=n^2+204n+10404=(n+102)^2$$ ... ? –  KM. Jun 3 at 18:33
    
@KM. The idea came looking for what I needed to add to $n^2+204n$ to get a perfect square. –  ajotatxe Jun 3 at 18:50
1  
@KM. Complete the square. Since $204n=2 \times 102 \times n$ , we are interested in $102^2=10404$ to make the expression a perfect square. This might be done to generate a number in the R.H.S. while making use of the factorization $k^2-m^2=(k+m)(k-m)$ and then applying elementary number theory. –  MathGod Jun 3 at 18:50

there must be some natural $k$ s.t. $n^2 + 204n = (n+k)^2$. easy to see the bigger the $k$ the bigger the $n$ because it's equivalent to $$ (204-2k)n = k^2$$ then again there's an easy bound on $k$ since $$n^2 + 204n < (n+102)^2$$ hence we can just try a couple of possibilities. if you write $$n^2 + 204n = (n+101)^2$$ then it won't give any solutions, since it boils down to $2n = 101^2$ on the other hand trying $$n^2 +204n = (n+100)^2$$ gives a solution $n=2500$ and we're done

share|improve this answer

So for some positive integer $m$, you must have $n^2+204n=m^2$. Considering this a quadratic in $n$, you need the discriminant $204^2+4m^2$ to be an even perfect square, say $4a^2$. So we have $a^2-m^2=10404 \implies (a+m)(a-m)=10404 \implies a+m = 5202, a-m=2$ as we want the largest $m$ (so as to get the largest $n$). Solving gives $m=2600 \implies n = 2500$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.