Pairs Satisfying a Cubic

$n^{3} + 2n^{2} = m^{2}$

$n^{2}(n+2)=m^{2}$

We know that for a product to be a square, each term must be a square.

$n^{2}$ is clearly a square, so $(n+2)$ must be a square.

$n+2=q^2$ and $1 \le n \le 1000$

The number of squares less that $1000$ is $31$ since $31^{2} < 1000 < 32^{2}.$

But we have to exlude $n + 2 = 1^{2}$ because $n = - 1$ and n must be a positive integer.

So the number of ordered pairs is : $31 - 1 = \boxed{30}$

$n^3+2n^2=m^2$ $\Leftrightarrow n^2(n+2)=m^2$ . Because $n; m$ are positive integers and $n^2(n+2)=m^2$ , we conclude that $n+2$ must be a perfect square $\Rightarrow n+2=k^2$ (k is a positive integer) But $1\le n\le 1000 \Rightarrow 2\le k\le 31$ . Hence, we have $30$ integers satisfying the equation which lead to $30$ pairs of $(n;m)$ satisfying the equation.

The pairs (n;m) are the pairs (n;squareroot of(n^3+2n^2) such that n is a positive integer from 1 to 1000 and "squareroot of(n^3+2n^2)" is also a positive ineger. (1) The number of pairs is the number of possible values n can take such that n is a positive integer from 1 to 1000 => for which squareroot of(n^3+2n^2) is a member of positive integers. which implicates that n^3+2n^2 is a perfect square. A=n^3+2n^2=n^2 (n+2) n^2 is a perfect square and we need A to be a perfect square, so let's find n such that n+2 is a perfect square. n+2 is a perfect square => we can find a positive integer x such that n+2=x^2 => n=x^2-2 n is a positive inter and 1≤n≤1000 => 1≤x^2-2≤1000 => 3≤x^2≤1003 => squareroot(3)≤x≤squareroot(1003) and since x is also is an integer => E(squareroot(3))+1≤x≤E(squareroot(1003)) => 2≤x≤31 so x is a member from the set I={2,3,4,......,31} so the number of possible values x can take are Card{I}=31-2+1=30 and since n=x^2-2 then the number of possible values n can take is the number of possible values x can take. => the the number of possible values n can take is 30 and building up from the proposition (1) We deduce that the number of pairs (m,n) is 30.

Since n^3+2n^2=n^2(n+2), n+2 must be a perfect square. And n+2 can be a perfect square 2^2,3^2,...,31^2 ; this is because n+2 can be an integer from 2 to 1002. So there are possible 30 pairs.

$n^3 + 2{n^2} = m^2$ ,then we factorize the $n$, so :

$n^2(n + 2) = m^2$ ,then move the $n^2$ to the right side, so :

$n + 2 = [\frac{m}{n}]^2$ $\sqrt[2]{(n + 2)} = \frac{m}{n}$

because $n$ is positive integers, then $\frac{m}{n}$ is positive integers. Because $1 ≤ n ≤ 1000$, and $\sqrt[2]{(n + 2)}$ is integers (positive integers), $n + 2$ max is $961 = 31^2$, so there are 31 solutions. but $1^2$ cannot be followed into the solutions because

$(n + 2) = 1^2$ $n + 2 = 1$ $n = -1$

so, there are 30 solutions of n and automatically there are 30 solutions of possible pairs of $(n,m)$.

n^2(n+2)=m^2 so, n+2=(m/n)^2 hence n+2 is a perfect square. find all such n's between 1 and 1000. there is no restriction on m, so dont worry about it.

$m=n(\sqrt{n+2})$ As m and n both are positive integers, it follows that $n+2$ is a perfect square. And also, $3\leq (n+2) \geq 1002$ $\sqrt{1002} > 31$ and $\sqrt{3} < 2$ Thus the various values which $n+2$ can take are $2^2, 3^2,........31^2$ So, n can take 30 different values and for each value of n, m has only one solution in positive integers. Therefore, number of solutions or pairs of(n,m) are 30.

$n^3 + 2n^2$ = $m^2$

$n^2$ can be factored out from left hand expression

$n^2( n + 2)$ = $m^2$

Taking square root of the whole equation : $n* sqrt(n + 2)$ = $m$

$m$ is an integer, therefore left hand side must also be an integer :

$sqrt(n + 2)$ is also an integer which means $n + 2$ is a square

Now, we only need to check number of squares below 1000 for every corresponding value of $n$ we will get $m$ .

We know 31's square is the largest square below 1000 Therefore answer is 30 ( excluding 1 : $n + 2$ cannot be equal to one { $n$ cannot <= 0})

I really liked this problem.

Factoring $n^3 + 2n^2$ to $\; n^2 \times (n+2)$ , we get that, for it to be a perfect square $m^2$ , $n+2$ must be a perfect square.

This will hold for $30$ values of $n$ , specifically $2^2 - 2, 3^2 - 2, \cdots 31^2 - 2$ , because $31^2 - 2$ is the greatest value of all that still is smaller than $1000$ (problem restriction).

by solving m= square root of {n n (n+2)} =n* square root of (n+2) n+2 must be perfect square , 1<=n<=1000 n+2 is =1,4,16,.........................961( square of 31)

first take the square root of the polynomial as we dont need to take care for the negative cases of m,n(given in the question). Now, we have n(n+2)=m. Now the last square possible is 31 square which can be contained in the under root. Hence from 2 square to 31 square we have 30 values of 'n' hence ans = 30 pairs for m,n as there in no restriction on 'm' except it being positive.

m = n * sqrt(n+2)

so smallest value of n is 2 and maximum is 31.

Hance total no. of values of n is 30.

$n^3 + 2n^2 = m^2$
$n^2(n + 2) = m^2$

$n^2$ and $m^2$ are always a square number. Then, $n+2$ must be a square number

Note : square number $\times$ square number = square number

$1\leq n\leq 1000$ $3\leq n+2 \leq 1002$

There are 30 square number between 2 and 1002 ( $4,9,16,25,36,\ldots,961$ )

Hence, there are $\boxed{30}$ pairs of $(n,m)$

function lookup(){ list=[]; for(n=1;n<1001;n++){ m=n*n*n+2*n*n; if(Math.floor(Math.sqrt(m))==Math.sqrt(m)){ list[list.length]=[n,m];} } return(list.length); }

The above expression can be neatly factored to $n^2(n + 2) = m^2$ . Since $n^2$ is a perfect square, $n+2$ must also be a perfect square. There are $\boxed{30}$ perfect squares (2 --> 31) for $n+2$ between 1 and 1000. (Note that 1 cannot work for n+2 because n would then have to be negative, thus out of the range for n)

Because n^3 + 2n^2 = m^2 , so we can simply the satisfy become n^2 ( n + 2 ) = m^2 --> So n that possible ordered is square number of n <= 1000 - 2 ---> n <= 998 ,except 1. So, the solution is root of 961 = 31 . The answer is 31 - 1 = 30.

Given,

$n^{3} + 2n^{2} = m^{2}$

$n^{2}(n + 2) = m^{2}$

$n \times \sqrt{n + 2} = m$

Here, the negative part of $m$ would not be considered as it is given $m$ is a positive integer.

Also, since $m$ is an integer, the term inside the square root must be a square, i.e. the term inside the square root must be 1, 4, 9, 16, 25 and so on.

Domain of $n$ is $1$ to $1000$ , so maximum the term inside the square root can go upto $31^{2} = 961$ .

It cannot be $32^{2} = 1024$ , as for that $n$ has to be 1022, which is not possible for the given domain of $n$ .

So, there are actually $31$ values of $n$ which will give positive integer values for $m$ .

But, there's a catch here. The term inside the bracket is $n + 2$ , which means it has to start with $2^{2} = 4$ . It cannot start with $1^{2} = 1$ because for that $n$ has to be $-1$ which is not possible, given the domain for $n$ .

So, actually there are $31 - 1 = \boxed{30}$ values of $n$ which gives positive integer values for $m$ .

Hence, number of ordered pairs of $(n, m)$ is $30$ .

That's the answer!

in order to "m" be a integer n^{3} + 2\times n^{2} will have to be a square number. factoring n^{3} + 2\times n^{2} we find n^{2} \times (n+2). as n^{2} is already square to make this number n has to be 2 less than a square number.there are distinct 31 square numbers between 1 to 1000. if n = 1 than m will not be integer. hence the answer is 30 . please help me write mathematical notation properly. tnx for time and attention.

$n^{3}$ + $2n^{2}$ = $m^{2}$

$n^{2}$ [ $n$ + 2 ] = $m^{2}$

by plugging in few values.

we get ( $n$ , $m$ ) as ( $2$ , $4$ ), ( $7$ , $21$ ), ( $14$ , $56$ ).

here we observe some awesome sequence 2, 7, 14, ….

after carefully observing we can conclude sequence is $n$ = $(x+1)^{2}$ - $2$ .

the $x+1$ because we would yield -1 after plugging $x$ = 1. and thats not in domain.

and … oh yeah the domain, 1 =< $n$ =< 1000 $$

$(x+1)^{2}$ - $2$ =< 1000

$(x+1)^{2}$ =< 1002

$x$ + $1$ =< $31.65$

$x$ =< $30.65$

$x$ = $\boxed{30}$

hence proved

n^3+2n^2=n^2(n+2) = m^2. From this we conclude that n + 2 must be a perfect square. Let it be k^2 Then n should be k^2-2. We will put k from 2 to 31 as at 32 n exceeds 10(putting k as 1 makes n = 0 which is not valid.)