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# NT Problem

How many perfect squares can there be of the form N=$$\overline{abba}$$ for some 4-digit positive integer N?

Note by Abhimanyu Swami
4 years, 3 months ago

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Note that we can write $$N=\overline{abba}$$ more mathematically rigorously as $$N=1000a+100b+10b+a=1001a+110b=11(91a+10b)$$. Since $$N$$ is a square, $$11|91a+10b$$.

Therefore we want to find all $$a,b$$ such that $$91a+10b\equiv 0\pmod{11}\implies 3a-b\equiv 0\pmod{11}$$.

Now we can do casework on $$a$$. Note that since the number of possible values of $$b$$ is less than the number of possible numbers modulo $$11$$, there must exist at most one unique solution to $$b$$ for every $$a$$, and there will not exist any solution for $$b$$ when $$3a\equiv 10\pmod{11}$$ which happens at $$a=7$$.

We list out the solutions: $$(a,b)=(1,3),(2,6),(3,9),(4,1),(5,4),(6,7),(8,2),(9,5)$$. Plugging this back in gives $$1331,2662,3993,4114,5445,6776,8228,9559$$. Factorizing gives that none of them are squares. Therefore there are $$\boxed{\textbf{no solutions}}$$

- 4 years, 3 months ago

Can you please explain what is meant by your line '.................the number of possible values of ..............happens at 7) with explaining a bit about modular arithmetic because I know very less about the topic.

- 4 years, 3 months ago

What I'm basically saying: we know that 3a-b has to be divisible by 11. Note that when we choose an "a", we might have some solutions for "b". However, notice that if we have a solution for "b", there can't be any more solutions because you can't have two single digit numbers both have the same remainder when divided by 11. Does that make sense?

- 4 years, 3 months ago

Correct, but be careful: this doesn't occur ONLY because "the number of possible values of $$b$$ is less than the number of possible numbers modulo $$11$$"; for example, if $$b$$ could only be the numbers $$5, 1, 16,$$ there are fewer possible values, but $$b$$ can be $$5$$ mod $$11$$ in two ways. What you said occurs mostly because the possible digits, $0, 1, 2, \ldots, 9,$ are consecutive, and there being only $$10$$ of them also contributes to this fact.

- 4 years, 3 months ago

Checking my memory for a list of 4 digit perfect squares, there aren't any of that form... but let's just make sure :P

$$N = 1001a + 110b = 11(91a + 10b)$$. So, $$11|91a + 10b \rightarrow 11|3a - b$$, but we also have to consider the other factors of $$91a + 10b$$, namely if they come in even pairs. I'll worry about later..

Since $$a, b < 10$$ and $$a > 0$$, the congruence $$3a - b \equiv 0 \pmod {11}$$ has very countable solutions in $$(1, 3), (2, 6), (3, 9), (4, 1), (5, 4), (6, 7), (8, 2), (9, 5)$$. These can be checked to not be perfect squares.

- 4 years, 3 months ago