Searching for the Perfect Square

$\frac{n}{1450-n} = \frac{1450}{1450-n}-1$ . $\frac{1450}{1450-n}$ must be a factor of 1450 to be a positive integer, so we see if any of the factors-1 is a perfect square. The factors are: 1, 2, 5, 10, 25, 29, 50, 58, 145, 290, 725,1450. We see that 1, 2, 5, 10, 50, 145, and 289 are the only ones that work, for a total of 7.

n/(1450 - n) = k^2

nk^2 - 1450k^2 + n = 0

n(k^2 + 1) - 1450k^2 = 0

n = 1450k^2/(k^2+1)

k^2 and k^2+1 always coprime

so, for n integer, 1450 should be divisible by k^2+1

we need to find out factors of 1450 that are in the form of k^2+1

1450 = 2 5^2 29

=> 1, 2, 5, 10, 25, 29, 50, 58, 145, 290, 725, 1450

k^2+1 = 1, 2, 5, 10, 50, 145, 290, satisfies

so 7 solutions for n also

Assume that $\frac{n}{1450-n}=k^2$ , where $k$ is an integer. Then $n=k^2(1450-n) \Rightarrow n+nk^2=1450k^2 \Rightarrow n=\frac{1450k^2}{k^2+1}$ . Since n is integer, so $k^2+1\mid1450k^2\Rightarrow k^2+1\mid 1450k^2-1450(k^2+1)=-1450$ . Hence, $k^2+1$ is a positive divisor of $-1450$ , which are $1,2,5,10,25,29,50,58,145,290,725,1450$ . But only $1,2,5,10,50,145,290$ can be written in the form $k^2+1$ where $k$ is integer, and for each value of $k^2$ we receive 1 value of $n$ . So the answer is 7.

If $\frac {n}{1450-n} = x^{2}$ then we can say that ${1450-n} \leq {n}$ .

$1450- n > 0$ so $n < 1450$

$n = x^2(1450-n)$

$1450x^2= n + x^2n$

$1450x^2 = n(x^2+1)$

$1450x^2 \leq 1449x^2 + 1449$

$x^2 \leq 1449$

$\sqrt{1449} \approx 38$ so $x \leq 38$

$n = \frac {1450x^2}{x^2+1}$

$\because n \in \mathbb{Z}, (x^2+1) \mid 1450x^2$

$1450x^2 \equiv 1450(x^2+1) - 1450$

$\therefore (x^2 + 1) \mid 1450x^2 \Rightarrow (x^2+1) \mid 1450$

$( x^2 + 1) \in \left \{1,2,5,10,25,29,50,58,145,290,725,1450\right \}$

$(x^2+1) > 0 \Rightarrow x^2 \in \left \{1,4,9,24,28,49,57,144,289,724,1449 \right \}$

$x \in \mathbb{Z} \Rightarrow x^2 \in \left \{0,1,4,9,49,144,289 \right \}$

$n_{0} = \frac {1450*0}{0+1} = 0$

$n_{1} = \frac {1450*1}{1+1} = 725$

$n_{2} = \frac {1450*4}{4+1} = 1160$

$n_{3} = \frac {1450*9}{9+1} = 1305$

$n_{4} = \frac {1450*49}{49+1} = 1421$

$n_{5} = \frac {1450*144}{144+1} = 1440$

$n_{6} = \frac {1450*289}{289+1} = 1445$

Hence there are 7 values of $n$ such that $\frac {n}{1450-n}$ is a perfect square.

n/(1450-n)=k^2
(k^2+1)n=1450 k^2
n=1450
k^2/(k^2+1).
Therefore an integral solution for n means having a factor of 1450 which is 1 more than a perfect square.
There are 7 such factors of 1450:
1, 2, 5, 10, 50, 145 and 290.
Thus, there are 7 integral values of n for which n/(1450-n) is a perfect square.

Let n/(1450-n)=k^2; n(1+k^2)=1450* k^2; n=1450 k^2/(1+k^2); 1450=2 5^2*29; So for n to be an integer, (1+k^2) has to be a factor of 1450. The 12 factors of 1450 are 1,2,5,10,25,29,50,58,145,290,725,1450. (1+k^2) has to be one of these 12 factors. But for k to be positive integer, only permissible values are k=0 gives n=0 k=1 gives n=725 k=2 gives n=1160 k=3 gives n=1305 k=7 gives n=1421 k=12 gives n=1440 k=17 gives n=1445 These are the only seven permissible values of n for which n/(1450-n) is a perfect square

Let $\frac {n}{1450-n}$ be equal to $x^2$ .

Expressing the equation $\frac {n}{1450-n}=x^2$ in terms of $n$ makes it $n=\frac {1450x^2}{x^2 + 1}$ .

Note that the positive factors of 1450 are 1, 2, 5, 10, 25, 29, 50, 58, 145, 290, 725 and 1450. Since $n$ is an integer, $1450x^2$ must be divisible by $x^2 + 1$ . This only happens when $x^2+1$ is a factor of 1450.

Therefore we only need to find the factors of 1450 that are one more than a perfect square. These are 1, 2, 5, 10, 50, 145 and 290. Therefore there are 7 possible values for $n$ , each of which is derived by substituting $x$ into the equation $n=\frac {1450x^2}{x^2 + 1}$ .

\frac {n}{1450-n} = \frac {1}{1450/n -1} ...(1) denominator of (1) needs to be equal to reciprocal of perfect squares i.e 1,\frac {1}{4},\frac {1}{9},... \Rightarrow \frac {1450}{n} -1 = 1,\frac {1}{4},\frac {1}{9},... \Rightarrow \frac {1450}{n} = 2,\frac {5}{4},\frac {10}{9},... \Rightarrow \frac {1450}{ 2,\frac {5}{4},\frac {10}{9},...} = n ( <,> denotes separate cases taken one at a time) now, number of distinct values of n = number of factors of 1450 of the form m^2 +1 which are 1,2,5,10,50,145,290 for m=0,1,2,3,7,12,17 hence, n has 7 such integer values.

At first, we see that our expression on the left must be a non-negative integer to be a square number. This discards the negatives values for n and values greater than $1450$ . $n = 1450$ gives a division by $0$ .

So we have $0 \leq n < 1450$ . $n = 0$ is a trivial solution.

With the equation $\frac{n}{1450-n} = a^2$ Testing the squares ${1, 4, 9, 16, 25, .., 144}$ for $a$ we see that $1, 4, 9, 49$ and $144$ gives a solution. For $n = 144$ we have $\frac{n}{1450-n} = a^2 \Rightarrow n = 1440$ .

Testing $n$ from $1441$ to $1449$ we see that only $1445$ gives a solution

Now, counting all the possibilities we have $n=0$ , $a =$ { $1, 4, 9, 49, 144$ }, $n = 1445$ with a total of $7$ wich is the answer.

as given

n/(1450-n) is a perfect square......so,

n/(1450-n)=x^2(suppose)

then cross multiply it

we get n x^2-1450 x^2+n=0.......from here n=(1450 x^2/(x^2+1))

now, 1450 should be divisible by x^2+1 then factor 1450 in the given form we get seven solutions 2,5,1,10,145,50,290............. therefore the answer is seven

Let the perfect square be $x^2$ , i.e. $n/(1450-n)=x^2$ . Simplifying, $\frac {1450 (x^2)}{(1+x^2)}=n$ . It follows that $n$ can be uniquely determined by $x$ . ( $n$ is a function of $x$ .) Since $x^2$ and $1+x^2$ are coprime, $n$ is an integer exactly when $\frac {1450}{(1+x^2)}$ is an integer. Since $1450=2\times 5 \times 5 \times 29$ , and hence has 12 positive factors, of which 7 can be expressed in the form $x^2+1$ for an integer $x$ , namely 1, 2, 5, 10, 50, 145 and 289. This gives 7 solutions for $x$ and hence $n$ .

[LaTeX edits - Calvin]

Solution 1: Let $\frac {n}{1450-n} = a^2$ , where $a$ is a non-negative integer. We first note that $n - 1450 \neq 0 \Rightarrow n \neq 1450$ . Rearranging for $n$ , we have $n = \frac{1450a^2}{1+a^2}$ . Since $\gcd(a^2,1+a^2)=1$ , thus $1 + a^2$ must divide $1450 = 2\times 5^2 \times 29$ . The divisors of 1450 are $1, 2, 5, 10, 25, 29, 50, 58, 145, 290, 725, 1450$ . The only values that have the form $1+a^2$ are $1, 2, 5, 10, 50, 145, 290$ , which correspond to $a= 0, 1, 2, 3, 7, 12, 17$ . Substituting in, this gives $n=0, 725, 1160, 1305, 1421, 1440, 1445$ , for a total of $7$ solutions.

Solution 2: We first rewrite \frac {n}{1450-n} = a^2) as $n=(1450-n)a^2$ , then $(1450-n)a^2-n=0,$ then $(1450-n)(a^2+1)=1450.$ So $1 + a^2$ must divide $1450 = 2\times 5^2 \times 29$ . The divisors of 1450 are $1, 2, 5, 10, 25, 29, 50, 58, 145, 290, 725, 1450$ . The only values that have the form $1+a^2$ are $1, 2, 5, 10, 50, 145, 290$ , which correspond to $a= 0, 1, 2, 3, 7, 12, 17$ . Substituting in, this gives $n=0, 725, 1160, 1305, 1421, 1440, 1445$ , for a total of $7$ solutions.