Simple Problem

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  Question: Find all two-digit positive integers such that the sum of the number and the number obtained by reversing the order of digits is a perfect square. Detailed solution: Let $xy$ be any such two-digit number (Note: Here $xy$ is just the form of a two-digit number like 25,43,etc., where $x$ and $y$ are digits). Therefore, $xy$ can be written as $10x+y$ (in expanded form). Number obtained by reversing the order of digits of $xy$ is $yx=10y+x$. According to question, The sum of $xy$ and the number obtained by reversing the order of digits of $xy$ i.e., $yx$ is a perfect square, which means $(10x+y)+(10y+x)=$Perfect square or, $11x+11y=$Perfect square or, $11(x+y)=$Perfect square .....$(1)$ But, maximum value of $x$ and $y$ are 9 each, so maximum value of $x+y$ is $9+9=18$. Observe that, to satisfy equation $(1), x+y=11k²$, for some positive integer $k$. For $k=1,x+y=11$ and for $k>1$ i.e., for $k≥2, x+y≥44$, which is not possible. Therefore, we get that $x+y=11$ is the only...
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A Question from Canadian Mathematical Olympiad 1971

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Question: Prove that the equation x³+11³=y³ has no solution in positive integers x and y.

Solution:

x³+11³=y³ =>y³–x³=11³ =>(y–x)(y²+yx+x²)=11³.

Case 1: y–x=1 and y²+yx+x²=11³
So, y=x+1 and (y–x)²+3yx=11³ => 1+3(x+1)x=1331 => 3(x+1)x=1330, which is not possible, because 3 does not divide 1330.

Case 2: y–x=11 and y²+yx+x²=11²
So, y=x+11 and (y–x)²+3yx=11² => 11²+3(x+11)x=11² => 3(x+11)x=0, which is not possible because this gives x(x+1)=0 => x=0 or x=–1, but neither of them is possible since x is a positive integer.

Case 3: y–x=11² and y²+yx+x²=11
So, y=x+11²and (y–x)²+3yx=11 => 11⁴+3(x+11²)x=11, which is not possible because this will give 3(x+11²)x=11–11⁴, which is negative.

Case 4: y–x=11³ and y²+yx+x²=1
So, y=x+11³ and (y–x)²+3yx=1 => 11⁶+3(x+11³)x=1, which is not possible because this will give 3(x+11³)x=1–11⁶, which is negative.
Since there is no possible case, so the equation x³+11³=y³ has no solutions in positive integers x and y.

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