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If we ${x}$ be the number that we want to square to obtain our 5-digit palindrome then it must satisfy: $111 \leq\ x < 316 \Rightarrow\ 12321 \leq\ x < 99856$ Now $x^{2}$ can end in a 1,4,5,6 or 9 , so let's start with 9 and go down if necessary. This means that $x^2 = 9aba9$ Keeping this in mind, let $\ x = pqr$ (with p = 3) such that: $(300+10q+r)^2 =90000 +1000(6q) +100(6r+q^2) +10(qr) +r^2$ It is clear that the palindrome must end with a 9, so $\ r = 3 \ or \ 7$

1) If $\ r = 3$ , then $x^{2} = 90000 + 1000(6q) +100(18+q^2) +10(6q) +9$ from which you can see that tens and thousands digit will not be equal because 18 + $q^{2}$ > 10 , so a digit will carry.

2) If $\ r = 7$ then: $x^2 = 90000 + 1000(6q) + 100(42+q^2) +10(14q) + 46$ Now by carrying factors of ten: $x^2 = 90000 +1000(6q+4) +100(2+q^2 +q) +10(4q+4) + 9$ If q= 0 then it is easy to see that $307^2 = \boxed{94249}$ There can be no higher numbers as $\ x$ must end in a 7 and $\ x < 316$