# BMO1 2011 Problem 1: Find $n$ such that $n^2+20n+11$ is a perfect square

1. Find all (positive or negative) integers $n$ for which

$$n^2+20n+11$$

is a perfect square. Remember that you must justify that you have found them all.

For the first part I did so:

$$n^2+20n+11=M^2$$ where M is an integer. That factors down to: $$(n+10)^2-M^2=89$$ $$(n+10-M)(n+10+M)=89$$ From there it is easy to find solutions as 89 is prime and it is a diophantine equation.

So the first part of the question is quite simple, however I am completely stumped on how to do the second part, how might we show that those are the only solutions?

• Maybe if you provide your solution to the first part it would be easier for us to help you with the second part. – RGS Nov 22 '16 at 13:49
• Actually you have done it. You have strarted from the equation and you have deduced from it that $n$ must be $35$ or $-55$. – ajotatxe Nov 22 '16 at 14:01
• But how do we justify that these are the only values of n? – user14513462563 Nov 22 '16 at 14:02
• @MishalMrinal What you and ajotatxe proved is that: if $n$ is such that $n^2+20n+11$ is a perfect square, then $n$ is necessarly either $35$ or $-55$, in other words: $n^2+20n+11\text{ is a perfect square}\Rightarrow n\in\{-55,35\}$. By contraposition: $n\neq 35 \land n\neq -55\Rightarrow n^2+20n+11\text{ isn't a perfect square}$. – Scientifica Nov 22 '16 at 14:10

We have $n^2+20n+11=(n+10)^2-89$. So our number is $89$ away from some perfect square. What do you know about the difference between consecutive squares? What does that say about the possibility of (not necessarily consecutive) squares that are $89$ away from one another?

• The difference between consecutive squares is $2n-1$. However i didn't quite get your second hint. – user14513462563 Nov 22 '16 at 13:59
• The difference between two squares is therefore a sum of consecutive odd numbers. Now you can just try out in what ways you can make a sum of consecutive odd numbers become $89$. There aren't too many possibilities. Or go with ajotatxe's answer, which I think is better. – Arthur Nov 22 '16 at 14:02
• I used the same approach as the answer below, however is doing the same thing outlined below enough of a justification that these are the only solutions for n? How do we confirm that we haven't missed any? – user14513462563 Nov 22 '16 at 14:04
• @MishalMrinal The difference is $2n+1$ ;) – Scientifica Nov 22 '16 at 14:07
• @Scientifica That depends on whether we have $n^2-(n-1)^2$ or $(n+1)^2-n^2$. Whether it's $+1$ or $-1$ is irrelevant in the end. The important point is that it is an odd number, and for the difference of the next pair of squares, it's the next odd number. So I'd say $2n-1$ is a good answer, and there is no need for you to correct it. – Arthur Nov 22 '16 at 14:10

Note that $$n^2+20n+11=(n+10)^2-89$$ so we have to find all the pairs of squares $a^2$, $b^2$ such that $a^2-b^2=89$. Of course, $a$ and $b$ are integers. We see that $|a|>|b|$.

Since $a^2-b^2=(a+b)(a-b)$ and $89$ is a prime number, then we have these possibilities:

• $a+b=89$ and $a-b=1$, that is, $a=45$, $b=44$.
• $a+b=1$ and $a-b=89$, that is, $a=45$, $b=-44$.
• $a+b=-89$ and $a-b=-1$, that is, $a=-45$, $b=-44$.
• $a+b=-1$ and $a-b=-89$, that is, $a=-45$, $b=44$.

So clearly, $a^2=2025$ and $b^2=1936$.

Then $n+10=\pm45$ which gives two values for $n$, namely $n=35$ and $n=-55$.

EDIT: How can be proved that there are no more solutions?

Indeed, this is already done. Our reasoning is like this:

• Proposition $A$: "$n$ is such that $n^2-20n+11$ is a perfect square".
• Proposition $B$: "$n$ is $35$ or $-55$".

We have shown, assuming that $n$ is integer, that $A$ implies $B$. So if $n$ is another integer number, $B$ is false. Hence, $A$ is false. That is, if $n$ is not $35$ or $-55$ then $n^2+20n+11$ is not a perfect square.

• I already did that part.....any help with the last part? – user14513462563 Nov 22 '16 at 14:01
• Read my comment to the question. – ajotatxe Nov 22 '16 at 14:02

Let $n^2+20n+11=(n+m)^2$

$\iff n=\dfrac{m^2-11}{2(10-m)}$

So, $m$ must be odd $=2r+1$(say)

$\implies n=\dfrac{2r^2+2r-5}{9-2r}$

Now if integer $d(>0)$ divides both $2r^2+2r-5,9-2r$

$d$ must divide $r(9-2r)+(2r^2+2r-5)=11r-5$

$d$ must divide $11(9-2r)+2(11r-5)=89$

As $n$ is an integer, $9-2r$ must divide $89$ whose factors are $\pm1,\pm89$