I have two equations, which may play an important role in my further studies on theory of numbers.

1) How many pairs of (A, B, x) we can make in $A^x + B ^x = prime$? Here $x$ is $> 2$ and A, B are positive integers.

2) can we find a number(s) with one hundred 0′s, one hundred 1′s and one hundred 2′s be a perfect square. If yes, what is that number or otherwise how to disprove it about such number does not existence?


  • $\begingroup$ clearly we must have $x=2^n$ or else the sum will be divisible by $A+B$ $\endgroup$ Dec 8 '13 at 13:12
  • $\begingroup$ clearly not. $2^2+3^2=14$ which is not divisible by 5 $\endgroup$
    – Thomas
    Dec 8 '13 at 13:17
  • $\begingroup$ @KonstantinosGaitanas and @Thomas! $1^4 + 2^4 = 17$. Here $x = 4 >2$ and A = 1, B = 2 are positive integers. The result 17 is prime. Like this how many A, B and x are existing? can you list? $\endgroup$
    – narosanair
    Dec 8 '13 at 13:20
  • $\begingroup$ @Thomas clearly yes.$x>2$ $\endgroup$ Dec 8 '13 at 13:21
  • $\begingroup$ Thank you for understanding me. can you list such pairs please... also look my second question of this post. $\endgroup$
    – narosanair
    Dec 8 '13 at 13:25

For the second question the answer is no.
The number you want is $222\cdots 111\cdots 000\cdots$ which is divisible by $3$ (because the sum of its digits is $2\cdot 100+1\cdot 100+0\cdot 100=300$ divisible by $3$)
But,it is not divisible by $9$ (because also the sum of its digits is not)
There for it must not be a square.

  • $\begingroup$ ! Sir, in my first question, you told that, x must be $2^n$ or else the sume will be divisible by $A + B$. Can you prove your argument mathematically? Please.. $\endgroup$
    – narosanair
    Dec 8 '13 at 14:45
  • $\begingroup$ If $x$ is not a power of $2$ it must have an odd factor say $d$. Then $x=k\cdot d$Then $A^x+B^x=(A^k)^d+(B^k)^d=(A^k+B^k)((A^k)^{d-1}-(A^k)^{d-2}\cdot B+\cdots +(B^k)^{d-1})$ which is a multiple of $A^k+B^k$.Don't you know the identity $a^n+b^n$ for odd $n$? $\endgroup$ Dec 8 '13 at 15:00
  • 1
    $\begingroup$ Note that although first he did not specify the order of the digits, this still works nicely :) $\endgroup$
    – chubakueno
    Dec 8 '13 at 18:38
  • $\begingroup$ @chubakueno I did not observe this,very sharp observation. $\endgroup$ Dec 8 '13 at 20:00
  • $\begingroup$ @Konstantioons Gaitanas! This is very good observation. Thanks a lot. $\endgroup$
    – narosanair
    Dec 9 '13 at 3:54

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