BdMO Online Forum

prove that, there exits no palindrome number of 2 digit which is a perfect square.no trial & error,plz.try to prove with logic.

all the two digit palindromes are $10k+k=11k$ type. where $k$ can be $1,2,3,4,5,6,7,8,9$ but if $11k=p^2$ $11$ must divide $k$. which it does not.so it isn't possible.

number 23. Reverse and add that number to 23 to yield the palindrome 55.

qeemat wrote:number 23. Reverse and add that number to 23 to yield the palindrome 55.

sorry,but i am a little confused what are you trying to say.
the condition says to find a two digit palindrome square.

The palindromes of two digits always have the same digit .So there are $9$ palindromes of two digit number. If there remains a perfect square palindrome,its first digit must be $1,4,5,6,9$ .So the palindromes must be $11,44,55,66,99$ .But neither of them is perfect square.So there doesn't exist any palindrome perfect square.

Prosenjit,
Your solution is close to brute force.you have to show that 11 must divide them.Eventually in order to be a perfect square 121 must divide them.Which is not possible.