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

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The equation \(4^a + 4a^2 + 4 = b^2\) has integer solutions for \(a\) and \(b\) only when \(a\) is less than or equal to 4. For \(a = 2\), the solution yields \(b = \pm 6\), while for \(a = 4\), \(b = 18\). The analysis shows that for \(a > 4\), no integer solutions exist due to the inequality \(4a^2 \geq 2^{a+2}\), which holds only for \(a \leq 4\). Thus, the integer pairs \((a, b)\) that satisfy the equation are limited to specific values derived from these calculations.

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Find no. of Integer value of $\left(a,b\right)$ which satisfy $4^a+4a^2+4 = b^2$
 
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jacks said:
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
 
Amer said:
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
 
try a = 4, b = 18
 
Last edited:
jacks said:
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 $a\leqslant 4.$
 

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