$-(1 + m)^2 - t - (-1 + r - w) (1 + r + w)$, solve for $w,m$ with known $t,r$?

Question

Consider some known fixed numbers $t,r$, then can we solve for $m,w$ in $-(1 + m)^2 - t - (-1 + r - w) (1 + r + w) = 0$ ?

Is there any way to figure out what are the integer solutions to this? I have tried looking at this every which way and nothing. I can manufacture many of this equations by choosing $t,r$ with known $w,m$ integers, but given $t,r$ only can we find $m,w$?

Update: $t,r,m,w \in \mathbb{N}$

Answer 1

You have $$-(1+m)^2-t-(-1+r-w)(1+r+w) = 0 $$ $$\Longleftrightarrow -(1+m)^2-t-(r-(w+1))(r+w+1) = 0 $$

$$\Longleftrightarrow -(1+m)^2-t-r^2+(w+1)^2 = 0 $$

$$\Longleftrightarrow (w-m)(w+m+2)=t+r^2 $$

Now if $t$ and $r$ are given, you can check the dividers of $t+r^2$ and deduce the different possibilities for $w-m$ and $w+m+2$, and then deduce the conrresponding values of $m$ and $w$.

Answer 2

"OP" equation is equal to:

$(m+1)^2=(w+r+1)(w-r+1)-(t)$

For given, $(r,t)=(1,47)$ we get:

$(m+1)^2=(w)(w+2)-(47)$

Solution is:

$w=p(7k^2-8k+9)$

$m=p(16k-5k^2-3)$

where, $p=[(1)/(k^2-1)]$

For, $k=2$, we get:

$(m,w,r,t)=(3,7,1,47)$

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