Find a pair of integers $n,x$ such that $84 = nx + (n-1)n$ and $x$ is odd

Question

I have a equation like this:

$$84 = nx + (n-1)n$$

where, $x$ is odd.

I need to find the fastest way to find a possible $n$ and $x$.

(In this case: $n = 6, x = 9$)

Edit: Maybe the background helps: 84 is the distance between 2 square numbers (100 - 16)

Answer 1

We have $84=n(x+n-1)$, hence $n\mid 84$. A possibility is to go through the divisors of $84$ and check if $x$ is an odd integer. It is already, for $n=2$.

Edit: We can find all solutions, but if you only want "a possible $n$ and $x$", this is the fastest.

Answer 2

As others have said, you have $84=n(n+x-1)$ and you are given that $x$ is odd, so $x-1$ is even and the two factors have the same parity. The factors can't both be odd, since that would give an odd product, so they must both be even.

$84=4\times 3 \times 7$, so you need a $2$ in each factor and split the $3$ and $7$ between them - there are four possibilities.

$84=2\times 42$ gives $n=2, x=41$

$84 = 6 \times 14$ gives $n=6, x= 9$

$84= 14 \times 6$ gives $n=14, x= -9$

$84 = 42 \times 2$ gives $n=42, x=-41$

Note you did not specify positive integers (integers seems to be implied, but is not explicit).

Answer 3

Solve for $x$ to get

$$x={84\over n}-(n-1)$$

so that $n$ must be a divisor of $84$. Note that if $n$ is odd, then $84/n$ and $n-1$ are both even, which would make $x$ even. So if we want $x$ to be odd, then we need $n=2m$, which gives us

$$x={42\over m}-(2m-1)$$

Since $2m-1$ is always odd, this expression gives an odd $x$ if and only if $42/m$ is even, which is to say, if and only if $m$ divides $21$. So there are eight solutions in all, corresponding to $n=\pm2$, $\pm6$, $\pm14$, and $\pm42$.

Remark: If you're only looking for positive values in $x$ and $n$, then it's easy to see from $x=(84/n)-(n-1)$ that $n$ must be less than $10$, so it suffices to try the divisors $n=1$, $2$, $3$, $4$, $6$, and $7$. The odd divisors are ruled out as before, and $n=4$ also gives an even value for $x$, so you're quickly left with $n=2$ and $6$, with $x=41$ and $9$, respectively.

Answer 4

We can read off a solution from your "background" that $\ 84 = \color{#0a0}{10}^2-\color{#c00}4^2.\,$ Completing the square

$$\ 4n(x\!+\!n\!-\!1)\, =\, (\color{#0a0}{x\!+\!2n\!-\!1})^2 - (\color{#c00}{x\!-\!1})^2 =\,4\cdot 84 = \color{#0a0}{20}^2 - \color{#c00}8^2\qquad\qquad\qquad\quad $$

The "obvious" solution $\ \color{#c00}{x\!-\!1=8}\,\Rightarrow\,x=9;\ \ \color{#0a0}{ x\!+\!2n\!-\!1 = 20}\,\Rightarrow\, n = 6\ $ is the one you seek.

If you aways have such "background", then this may be the simple method that you seek.