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

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Discussion Overview

The discussion revolves around finding integer solutions for the equation \(4^a + 4a^2 + 4 = b^2\). Participants explore specific integer values for \(a\) and \(b\) and examine conditions under which solutions may or may not exist, focusing on both theoretical and mathematical reasoning.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests testing \(a = 2\) and finds that \(b = \pm 3\), but another later corrects this to \(b = \pm 6\).
  • Another participant proposes testing \(a = 4\) with \(b = 18\) but does not provide further elaboration on this case.
  • A participant argues that there are no solutions for \(a > 4\) by analyzing the equation and deriving conditions that lead to the conclusion that \(a\) must be less than or equal to 4.
  • This participant notes that \(b\) must be even and discusses the implications of this on the equation, leading to a comparison of terms that suggests a limit on \(a\).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the integer solutions, as there are differing claims about specific values of \(b\) for given \(a\) and ongoing debate about the existence of solutions for \(a > 4.

Contextual Notes

The discussion includes assumptions about the parity of \(b\) and the implications of the derived inequalities, but these assumptions are not universally accepted or resolved among participants.

juantheron
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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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