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- Thread starter jacks
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- #1

- Mar 31, 2013

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The no. of positive Integer ordered pair $(a,b)$ in $4^a+4a^2+4 = b^2$

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

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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

- Mar 31, 2013

- 1,341

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

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

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

- Mar 31, 2013

- 1,341

A slight different solution