Integer Solutions for $4^a+4a^2+4=b^2$

[juantheron]

Find no. of Integer value of $\left(a,b\right)$ which satisfy $4^a+4a^2+4 = b^2$

 

[Amer]

Find no. of Integer value of $\left(a,b\right)$ which satisfy $4^a+4a^2+4 = b^2$

take a =2

$$4^2 + 4(2)^2 + 4 = 4(4 + 4 + 1) = 4(9)$$
$$b^2 = 36 \Rightarrow b = \pm 3$$

 

[topsquark]

take a =2

$$4^2 + 4(2)^2 + 4 = 4(4 + 4 + 1) = 4(9)$$
$$b^2 = 36 \Rightarrow b = \pm 3$$

Actually $$b = \pm 6$$

-Dan

 

[Amer]

try a = 4, b = 18

 

[Opalg]

Find no. of Integer value of $\left(a,b\right)$ which satisfy $4^a+4a^2+4 = b^2$

To see that there are no solutions with $a>4$, notice first that $b$ must be even. Next, the equation $4^a+4a^2+4 = b^2$ can be written as $(2^a)^2+4a^2+4 = b^2$. Since $2^a$ is even, and $b$ is also even, the smallest possible value for $b$ would be $2^a+2$. But $(2^a+2)^2 = 4^a + 2^{a+2} + 4.$ Therefore $$b^2 = 4^a+4a^2+4 \geqslant 4^a + 2^{a+2} + 4 ,$$ from which $4a^2 \geqslant 2^{a+2}$ and hence $a^2\geqslant 2^a.$ That only happens when $a\leqslant 4.$