National Number Theory

For discussing Olympiad Level Number Theory problems
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National Number Theory

Unread post by Kazi_Zareer » Tue Jan 05, 2016 2:26 pm

সকল মৌলিক সংখ্যা $p$ এবং স্বাভাবিক সংখ্যা $a,b$ বের কর যেন $p^a + p^b $ একটি square number হয়।

Find all prime number $p$ and natural number $a,b$, such $p^a + p^b $ is a square number.
We cannot solve our problems with the same thinking we used when we create them.

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Re: National Number Theory

Unread post by tanmoy » Wed Jan 06, 2016 12:28 pm

If $a=b$,then $2p^{a}$ is a perfect square.So,$p=2$ and $a$ is odd.
Now,suppose $a<b$.So,$p^{a}+p^{b}=x^{2}$ for some $x$.Or,$p^{a}(1+p^{b-a})=x^{2}$.$gcd(p^{a},1+p^{b-a})=1$,so,$p^{a}$ and $1+p^{b-a}$ both are perfect squares.$\therefore$ $a$ is even.Now,let $1+p^{b-a}=y^{2}$ for some $y$.Or,$p^{b-a}=(y+1)(y-1)$.As $p$ is a prime number,both $(y+1)$ and $(y-1)$ are power of $p$.Let $(y+1)=p^{m}$ and $(y-1)=p^{n}$.So,$p^{m}-p^{n}=2$.Or,$p^{n}(p^{m-n}-1)=2$.So,either $p^{n}=1$ or $p^{n}=2$.

If $p^{n}=1$,then $y=2$,$p^{m}=3$ which implies that $p=3$.So,$3^{b-a}=3$.$\therefore$ $b-a=1$,or $b=a+1$.

Now,if $p^{n}=2$,then $p=2$,$y=3$,so,$2^{b-a}=8$,so,$b-a=3$.Or,$b=a+3$.

So,the solutions are:$p=2$ and $a=b$=an odd positive integer;$p=2$ and $b=a+3$ where $a$ is an even positive integer;$p=3$ and $b=a+1$ where $a$ is an even positive integer.
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