Exploring Integer Ordered Pairs in a Unique Equation

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Discussion Overview

The discussion revolves around finding the number of positive integer ordered pairs \((a,b)\) that satisfy the equation \(4^a + 4a^2 + 4 = b^2\). The scope includes mathematical reasoning and exploration of potential solutions.

Discussion Character

  • Mathematical reasoning, Exploratory

Main Points Raised

  • Some participants propose that the equation can be analyzed by rewriting \(4^a\) as \(2^{2a}\) and exploring inequalities involving \(4^a + 4a^2 + 4\) and \((2^a + 1)^2\).
  • One participant suggests that if \(4^a + 4a^2 + 4 \geq (2^a + 1)^2\), it leads to the condition \(2 \cdot 2^a + 1 \leq 4a^2 + 4\), which implies \(a \leq 6\) and necessitates checking values of \(a\) from 1 to 6.
  • Another participant reiterates the approach of comparing \(4^a + 4a^2 + 4\) with \((2^a + 1)^2\) and emphasizes that if the former is less than the latter, a perfect square cannot exist.
  • A participant mentions a slightly different solution approach but does not elaborate on the specifics.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and propose different methods for approaching the problem, indicating that there is no consensus on a singular solution or method.

Contextual Notes

The discussion includes assumptions about the relationships between the terms in the equation and the conditions under which solutions may exist, but these assumptions are not universally accepted or verified.

juantheron
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The no. of positive Integer ordered pair $(a,b)$ in $4^a+4a^2+4 = b^2$
 
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jacks said:
The no. of positive Integer ordered pair $(a,b)$ in $4^a+4a^2+4 = b^2$
it may have a solution as 4^a = 2^ 2a

if

(4^a + 4a^2 + 4) >= (2^a+1)^2

or 2 * 2^a + 1 <= 4a^2 + 4

this gives a <= 6 and we need to check for a = 1 to 6.
rest is trivial
 
To kaliprasad

I did not understand it

it may have a solution as $4^a = 2^ {2a}$

if $\left(4^a + 4a^2 + 4 \right) >= (2^a+1)^2$

or $2 \cdot 2^a + 1 \leq 4a^2 + 4$

this gives $a\leq 6$ and we need to check for $a =1$ to $a = 6$
.
rest is trivial

would you like to explain me.

Thanks
 
jacks said:
To kaliprasad

I did not understand it

it may have a solution as $4^a = 2^ {2a}$

if $\left(4^a + 4a^2 + 4 \right) >= (2^a+1)^2$

or $2 \cdot 2^a + 1 \leq 4a^2 + 4$

this gives $a\leq 6$ and we need to check for $a =1$ to $a = 6$
.
rest is trivial

would you like to explain me.

Thanks

you know 4^a = (2^2a) = (2^a)^2

so square root is 2^a

next square has to be (2^a+ 1)^2

if 4^a + 4a^2 + 4 < (2^a+1)^2 we cannot have a pefect squre because we are less than the next square
 

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