Some Last year Divisional Problems

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Phlembac Adib Hasan
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Re: Some Last year Divisional Problems

Unread post by Phlembac Adib Hasan » Fri Jan 25, 2013 10:11 pm

skb wrote:Problem 1 - I solved it in a lengthy (also complex) way, I guess
Suppose, $a < b$ and $b = a+m$
and $p^a + p^b = k^2$
or, $p^a + p^{a+m} = k^2$
or, $p^a (1 + p^m) = k^2$
Now, we have two subcases, either p is even or odd
Subcase 1 - When $p$ is even, $(p=2)$
$1 + 2^m$ cannot be divided by $2$ and there should be pairs of integers in the dividers of a perfect square
so, both $2^a$ and $(1 + 2^m)$ are perfect squares
let, $1 + 2^m = n^2$
so, $2^m = (n+1)(n-1)$
as, $2^m$ cannot be divided by any other prime, so $n+1$ and $n-1$ both are power of $2$
it is possible only if $n=3$
so, here, $p=2$ , $a =\; $ any even number, $b = a+3$

Subcase -2 When $p$ is odd
using the same logic,
$p^a$ and $(1 + p^m)$ both are perfect squares
$1 + p^m = f^2$
$p^m = (f+1)(f-1)$
again, $f+1$ and $f-1$ both are power of $p$
but it is not possible in any case
so here should be no answer

(I am confused whether I'm right or wrong) :!: :!:
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sakib.creza
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Re: Some Last year Divisional Problems

Unread post by sakib.creza » Sat Jan 26, 2013 9:16 pm

skb wrote: Subcase 1 - When $p$ is even, $(p=2)$
$1 + 2^m$ cannot be divided by $2$ and there should be pairs of integers in the dividers of a perfect square
so, both $2^a$ and $(1 + 2^m)$ are perfect squares
এই লাইন টা বুঝলাম না
Last edited by *Mahi* on Sat Jan 26, 2013 10:48 pm, edited 1 time in total.
Reason: Transliterated

skb
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Re: Some Last year Divisional Problems

Unread post by skb » Sat Jan 26, 2013 10:23 pm

when $p=2$, $p^a(1+p^m) = 2^a(1+2^m)$
here, $1 + 2^m$ is odd and $2^a$ is even
as their product is a perfect square, and $1 + 2^m$ cannot be divided by $2^a$
so, both of them are perfect square separately
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Shadman95
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Re: Some Last year Divisional Problems

Unread post by Shadman95 » Tue Jan 29, 2013 3:19 pm

I guess, the answer of first one will be 2 to the power 4028 ... Can anyone ensure me?? :?:

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