# How many $4$-digit perfect square numbers are there whose last two digits are identical?

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!!!

• I mean, trial and error is pretty easy.
– lulu
Aug 11 '21 at 17:26
• 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. Aug 11 '21 at 17:29
• @cos_dm_math21 Note that $99$ can't be the ending since such numbers would be $3\pmod 4$.
– lulu
Aug 11 '21 at 17:31
• @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? Aug 11 '21 at 17:33
• @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. Aug 11 '21 at 17:36

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 $$38, 62, 88$$.

• You're missing a lot of solutions Aug 11 '21 at 17:37
• @egglog Can you give any example of a solution I'm missing? Aug 11 '21 at 17:37
• @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". Aug 11 '21 at 17:38