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Let $N$ be a two digit number and let $M$ be the number formed from $M$ by reversing $N$'s digits. The digits of $M^2$ are precisely those of $N^2$, but reversed.

$Proof$:

Since $N$ is a two digit number, we can write $N = 10a + b$ where $a$ and $b$ are the digits of $N$. Since $M$ is formed from $N$ by reversing digits, $M = 10b + a$.

$N^2 = (10a + b)^2 = 100a^2 + 20ab + b^2 $. The digits of $N^2$ are $a^2, 2ab, b^2$.

$M^2 = (10b + a)^2 = 100b^2 + 20ab + a^2$. The digits of $M^2$ are $b^2, 2ab, a^2$, exactly the reverse of $N^2$.

This proposition is false. Let $N$ be $15$. That means the proof above is not correct, but I can't see where exactly.

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    $\begingroup$ Carries? Those values aren't usually going to be a single digit. $\endgroup$ – Mike Jul 6 '14 at 17:44
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    $\begingroup$ $b$ is a single digit, but $b^2$ might not be. So the units digit of $N^2$ is not $b^2$; it is $b^2\bmod{10}$. Your other claims about the digits of $M^2$ and $N^2$ are similarly wrong. $\endgroup$ – MJD Jul 6 '14 at 17:44
  • $\begingroup$ So, if $a = 7$, then the $N^2$'s digit in hundredth place would be 49 which is not a digit. Is this one of the reasons this proofs wrong? $\endgroup$ – Prostitute Jul 6 '14 at 17:52
  • $\begingroup$ The only numbers for which the statement is true, are 11,22,33,44,55,66,77,88, and 99. $\endgroup$ – Klaas van Aarsen Jul 6 '14 at 18:08
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    $\begingroup$ No, it's true for 11 and 22 but false for 33 through 99. It's also true for 12, 13, 21, and 31. And even 10, 20, and 30 if suitably interpreted. $\endgroup$ – Greg Martin Jul 6 '14 at 18:44
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As Greg's examples and other comments point out this, can only be true if $a^2,2ab$ and $b^2$ are all less than $10$. Otherwise there is a carryover that spoils it, as your example shows...

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however there exists an infinite sequence of such numbers: we have

$12^2=144$

$102^2=10404$

$1002^2=1004004$

$10002^2=100040004$

$\dots\ \dots\ \dots$

and the same with digits reversed:

$21^2=441$

$201^2=40401$

$2001^2=4004001$

$20001^2=400040001$

$\dots\ \dots\ \dots$

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  • $\begingroup$ I don't think this addresses the actual question. $\endgroup$ – punctured dusk Mar 10 '15 at 20:04
  • $\begingroup$ Also, I remember reading on meta (I can't find the link) that there is nothing wrong with copying earlier answers you gave, but it is encouraged to add a link to the other answer. Here is it: math.stackexchange.com/a/1184224/43288 $\endgroup$ – punctured dusk Mar 10 '15 at 20:15
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It is too late to answer but I asked a similar question .so I am starting from your steps.You are using the assumption that square of $N$ (and $M$) are $\textit{three digits}$. Let $N^{2}=(10a+b)^{2}=100x+10y+z$ where $x,y,z$ are non negative integers and $x,z \neq 0$. Since $M^{2}$ is the reverse of $N^{2}$ then $M^{2}= 100z+10y+x$

If you are using one more assumption that $a>b$ then $N^{2}-M^{2}$ will lead to $a^{2}-b^{2} \leq 8$ because $x,z \neq 0$.

Now by substituting values foe $b$ and solving we get values for $a$.Then you can find result stated in starting of your problem is not even valid for all two digit numbers less then $33$ . Actually it holds good for only 21 and 31.If we use digits need not be distinct then result is very simple.

But I am not sure about two digit numbers whose square is a four digit number

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