প্রতিবার ম্যাথমেটিক্যাল কিছু লেখার সময় বোল্ড ফন্ট ইউজ করেছ। সুন্দর, চমৎকার, উত্তম। তবে প্রত্যেকটা $$ এবং $$ র স্থানে \$ ব্যবহার করলে আরও ভালো লাগত। তোমার প্রতিটা পোস্টে আমি যা করেছি সেটা হচ্ছে প্রতিটা $,$ মুছে তার জায়গায় ডলার বসিয়েছি। এতেই ল্যাটেক হয়ে গেছে। তবে এটা বড়ই পেইনফুল এবং বোরিং কাজ। আশা করি ভবিষ্যতে আর এটা করতে হবে না। আর sakib.creza, তোমার প্রতি একই অনুরোধ।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)
Some Last year Divisional Problems
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Re: Some Last year Divisional Problems
এই লাইন টা বুঝলাম নাskb 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
Last edited by *Mahi* on Sat Jan 26, 2013 10:48 pm, edited 1 time in total.
Reason: Transliterated
Reason: Transliterated
Re: Some Last year Divisional Problems
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
and you can download avro from here- http://www.omicronlab.com/avro-keyboard-download.html in order to use Bangla
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
and you can download avro from here- http://www.omicronlab.com/avro-keyboard-download.html in order to use Bangla
Dream shall never die.
Re: Some Last year Divisional Problems
I guess, the answer of first one will be 2 to the power 4028 ... Can anyone ensure me??