If $a(\cos x-1)+b^2=\cos(ax+b^2)-1$ hold true for all $x\in \mathbb{R}$, Then $(a,b)$ 
Number of positive integer ordered pair $(a,b)$ for which the equation
$a(\cos x-1)+b^2=\cos(ax+b^2)-1$ hold true for all $x\in \mathbb{R}$

$\bf{My\; Try::}$ We can write above equation as $a(\cos x-1)+b^2+1 = \cos(ax+b^2)$
Now Using $\cos x\leq 1\Rightarrow a(\cos x-1)+b^2+1 \leq b^2+1$
and $-1 \leq \cos (ax+b^2)\leq 1$
Now How can I solve after that, Help me
Thanks

$x=0$ gives $\cos(b^2)=b^2+1$. However $b^2+1\geq 1 \geq \cos(x)$, with equality only if $b=0$.
Since your question requires positive integers, we're technically done. 
However, I shall proceed to find all integers $a$ for which it holds.
Now we get $a(\cos(x)-1)=\cos(ax)-1$. $x=\frac12 \pi$ gives $-a=\cos(a\frac12 \pi)-1$.
But $a$ is an integer, so $\cos(a\frac12 \pi)$ is -1,0 or 1. so $-a=-1$, $-a=-2$ or $-a=0$, so $a=0$, $a=1$ or $a=2$. $a=0$ and $a=1$ indeed work.
For $a=2$ let $x=\pi$ and we get $2\cdot(-2)=\cos(2\pi)-1=0$, which is not true.
One can also prove that this are in fact all real solutions.

Conclusion: For positive integers $(a,b)$ there exists no such pair.
In the real numbers, the only pairs are $(0,0)$ and $(1,0)$.
