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From trial and error, I found only $6$ numbers i.e. $1600,2500,3600,4900,6400$ and $8100$ but in the answer for this question it has been given that there are $9$ such numbers. Which numbers I missed? Is there any mathematical way to solve this problem? Please help me on this!!!

Thanks in advance !!!

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    $\begingroup$ I mean, trial and error is pretty easy. $\endgroup$
    – lulu
    Aug 11 '21 at 17:26
  • $\begingroup$ For sure the last 2 digits can't be 1, 2, 3, 5, 6, 7, 8 or 9. So try to find the numbers of the form $\overline{ab44}$ which are perfect squares. $\endgroup$ Aug 11 '21 at 17:29
  • $\begingroup$ @cos_dm_math21 Note that $99$ can't be the ending since such numbers would be $3\pmod 4$. $\endgroup$
    – lulu
    Aug 11 '21 at 17:31
  • $\begingroup$ @cos_dm_math21 : I am sorry but I couldn't understand why the last digit needs to be 4 or 9 to be repetitive at the last two places? $\endgroup$
    – Ganit
    Aug 11 '21 at 17:33
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    $\begingroup$ @Ganit Try to argue modulo 3, 4 or 5. For example, a perfect square is not $2$ or $3$ $\pmod 5$. So you can get rid of 2, 3, 7 and 8. $\endgroup$ Aug 11 '21 at 17:36
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Writing some quick Python:

for x in range(100):
    y = x*x;
    if 1000 <= y and y < 10000 and (y%100)//10 == (y%10):
        print(y)

gives the following perfect squares:

1444
1600
2500
3600
3844
4900
6400
7744
8100

For a more theoretical approach, let's consider the squares $\mod 10$, which are $0, 1, 4, 5, 6, 9$.

So we need to check whether the numbers 0, 11, 44, 55, 66, and 99 are squares $\mod 100$. It's easy to see that $0 = 0^2$ and $44 = 12^2 = (100 - 12)^2$ are squares mod 100. It turns out the others are not squares mod 100.

Why can't $11, 55, 66, 99$ be squares? Because they are all either $3$ or $2$ mod 4, and $3$ and $2$ are not squares mod 4.

Now what numbers squared give us an ending of $00$? Such numbers would need to be multiples of 10, since their square is a multiple of 100.

What numbers squared give us $44$? These clearly need to end with either a $2$ or an $8$. Let's say the number is $10x + 2$ for $0 \leq x < 9$. Then we have $100x^2 + 40x + 4 = 40x + 4 = 44$. Then $40(x - 1) = 0$, so $x = 1$ or $x = 6$. So we're looking for the squares of numbers ending in $12$ or $62$.

A similar analysis for numbers ending in 8 means we're looking for the squares of numbers ending in 88 or 38.

So we need to get the squares of numbers ending in $x0$, $88$, $38$, $12$, or $62$. Now $31^2 < 1000 < 32^2$ and $10000 = 100^2$, so we're looking for the squares of numbers between $32$ and $99$ which end in $x0$, $88$, $38$, $12$, or $62$. The numbers of the form $x0$ are $40, 50, 60, 70, 80, 90$, which gives us 6 solutions. And the other solutions are the squares of $38, 62, 88$.

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  • $\begingroup$ You're missing a lot of solutions $\endgroup$
    – egglog
    Aug 11 '21 at 17:37
  • $\begingroup$ @egglog Can you give any example of a solution I'm missing? $\endgroup$ Aug 11 '21 at 17:37
  • $\begingroup$ @egglog This is the correct answer if "identical last two digits" means "identical to each other". Yours is correct if it means "identical to those in some other square". $\endgroup$ Aug 11 '21 at 17:38

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