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 1969

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Question: Prove that there are no integers a, b and c for which a²+b²–8c=6

Detailed solution: a²+b²–8c=6 gives a²+b²=6+8c.
Since, the RHS is even, so LHS is even, which means a²+b² is even. So we get two cases in which the first one is that a² and b² both are even and in the second one, a² and b² both are odd.

Case 1: a² and b² are both even. This gives that a and b both are even.
So, a=2p and b=2q for some integers p and q.
Substituting the values of a and b in the equation a²+b²–8c=6, we get
(2p)²+(2q)²–8c=6
or, 4p²+4q²–8c=6
or, 2p²+2q²–4c=3 (Dividing both sides by 2)
Clearly, the LHS is even and RHS is odd, which is not not possible. So we get a contradiction in this case.

Case 2: a² and b² both are odd. This gives that a and b are both odd.
So, a=2m+1 and b=2n+1 for some integers m and n.
Substituting the values of a and b in the equation a²+b²–8c=6, we get
(2m+1)²+(2n+1)²–8c=6
or, 4m²+4m+1+4n²+4n+1–8c=6
or, 4m(m+1)+4n(n+1)=8c+4
or, m(m+1)+n(n+1)=2c+1   ....(1) (Dividing both sides by 4)
If m is odd, m+1 is even and if m is even, m+1 is odd.
So, m(m+1) is always even and similarly, n(n+1) is always even.
Therefore, in equation (1), LHS is even and RHS is clearly odd.

Hence,  both the cases give contradictions. This means that there are no integers a, b and c for which a²+b²–8c=6.

Thank you... 



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