Thread: Perfect squares of n and n+99

1. Perfect squares of n and n+99

How many positive integer solutions are there n such that both n and n+99 are perfect squares?

I believe there is only 1 solution.

Any tips on how to tackle this? I haven't really come up with a method or a formula to calculate it, simply gone through the first 20 squares, and I doubt there will be anything after that

2. i tried the first couple on paper only the number 1 is working. so i guess that it

3. Originally Posted by Mukilab
How many positive integer solutions are there n such that both n and n+99 are perfect squares?

I believe there is only 1 solution.

Any tips on how to tackle this? I haven't really come up with a method or a formula to calculate it, simply gone through the first 20 squares, and I doubt there will be anything after that

I can't provide you with an equation or algorithm but I used an Excel table and have found:$\displaystyle \begin{array}{cccc}nr&sqr&NR&SQR\\1& 1& 10& 100\\ 15& 225& 18& 324\\ 49& 2401& 50& 2500\end{array}$

Now the difference between the squared numbers is 1. So there can't be any other pair of numbers which satisfy the given conditions.

4. How does 49 satisfy the conditions?

You must have misread my question

take n as 1
1^2 is a perfect square

(1+99)^2 = 100 = 10^2

49 does not suit this

49+99=148

15 does not suit this

5. Originally Posted by earboth
I can't provide you with an equation or algorithm but I used an Excel table and have found:$\displaystyle \begin{array}{cccc}nr&sqr&NR&SQR\\1& 1& 10& 100\\ 15& 225& 18& 324\\ 49& 2401& 50& 2500\end{array}$

Now the difference between the squared numbers is 1. So there can't be any other pair of numbers which satisfy the given conditions.
$\displaystyle n \in \{1, 225, 2401\}$ is correct. The answer can be obtained with a Diophantine equation.

$\displaystyle n = k^2$

$\displaystyle n+99 = m^2$

$\displaystyle k^2-m^2+99 = 0$

See Dario Alpern's quadratic Diophantine solver and plug in $\displaystyle a=1, c=-1, f=99$ and select "step-by-step".

Edit: Actually, the steps shown for this particular problem aren't as illuminating as they often are. Not sure the best way to work it out on paper.

Edit 2: See Soroban's post.. seems so obvious now.

6. Originally Posted by Mukilab
How does 49 satisfy the conditions?

You must have misread my question

take n as 1
1^2 is a perfect square

(1+99)^2 = 100 = 10^2

49 does not suit this

49+99=148

15 does not suit this

So take

$\displaystyle 225 +99 = 324 = 18^2$

$\displaystyle 2401+99 = 2500 = 50^2$

7. that's taking it as n^2+99 but the actual question is n+99

8. By the way, I worked it out by brute force initially as well using PARI/GP which is a great tool (and hopefully you won't get spoiled by it). I wrote

Code:
f(n)=if(sqrt(n)==floor(sqrt(n)),1,0)
for(i=0,10000,if(f(i)&&f(i+99),print(i)))
Could also have used

Code:
f(n)=if(floor(sqrt(n))^2==n,1,0)
which ought to be a bit more efficient.

9. Originally Posted by Mukilab
that's taking it as n^2+99 but the actual question is n+99
There are three solutions.

$\displaystyle n=1=1^2$ where $\displaystyle n+99=100=10^2$

$\displaystyle n=225=15^2$ where $\displaystyle n+99=324=18^2$

$\displaystyle n=2401=49^2$ where $\displaystyle n+99=2500=50^2$

10. Hello, Mukilab!

How many positive integers $\displaystyle n$ are there
such that both $\displaystyle n \text{ and }n+99$ are perfect squares?

We have two squares that differ by 99: .$\displaystyle a^2 - b^2 \:=\:99$

So we have: .$\displaystyle (a+b)(a-b) \:=\:\begin{Bmatrix}99\cdot1 \\ 33\cdot 3 \\ 11\cdot 9 \end{Bmatrix}$

Solve the three systems of equations:

. . $\displaystyle \begin{array}{ccc}a+b &=& 99 \\ a-b &=& 1 \end{array} \quad\Rightarrow\quad (a,b) \:=\:(50,49) \quad\Rightarrow\quad n = 2401$

. . $\displaystyle \begin{array}{ccc}a+b &=& 33 \\ a-b &=& 3\end{array} \quad\Rightarrow\quad (a,b) = (18,15) \quad\Rightarrow\quad n = 225$

. . $\displaystyle \begin{array}{ccc}a+b &=& 11 \\ a-b &=& 9 \end{array} \quad\Rightarrow\quad (a,b) = (10,1) \quad\Rightarrow\quad n = 1$