61 Given that $\left(x^{n}+c\right)^{m}$ is identically equal to $\left(a x^{m}+1\right)\left(b x^{m}+1\right)$ (where $m, n$ are both integers), then the value of $|a+b+c|$ is . $\qquad$
613 or 2. Since $\left(x^{n}+c\right)^{m} \equiv\left(a x^{m}+1\right)\left(b x^{m}+1\right)$, let $x=0$, we get $c^{m}=1$. If $m=1$, then $x^{n}+1 \equiv a b x^{2}+(a+b) x+1$. Thus, $n=2, a b=1, a+b=0$, which is impossible; or $n=1, a b=0, a+b=1$, at this time $|a+b+c|=2$. If $m \neq 1$, then the highest degree of the term in $\left(x^{n}+c\right)^{m}$ is $n m$, and the highest degree of the term in $\left(a x^{m}+1\right)\left(b x^{m}+1\right)$ is $2 m$, so $n=2$, and $\left(a x^{m}+1\right)\left(b x^{m}+1\right)=a b x^{2 m}+(a+b) x^{m}+1$, meaning $\left(x^{n}+c\right)^{m}$ has only three terms, thus we know $m=2$. Therefore, $\left(x^{2} \pm 1\right)^{2} \cong a b x^{4}+(a+b) x^{2}+1$. Hence $a b=1, a+b= \pm 2$, at this time $|a+b+c|=3$.