A four-digit number (not starting in zero) has the following properties:
What is this number?
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That's an interesting property. Among all 5-digit integers, only $81225$ has the following properties:
(i) it is a perfect square,
(ii) its first two digits are a perfect square, and
(ii) its last three digits a perfect square not ending (or beginning) with a zero.
If we remove the "not ending (or beginning) with a zero" we would have to include $16900, 36100$ and $64009.$ There are no 5-digit perfect squares where both the first three digits and last two digits are perfect squares of positive integers.
Among 6-digit numbers $225625$ is the only one which meets comparable conditions to those listed above, (with (ii) being altered to the first three digits being a perfect square). Relaxing the "zero condition" would allow for the inclusion of $144400, 256036, 324900$ and $576081$ .
Its curious that we have $1681, 81225, 225625$ as the first three numbers in this "special" sequence, (noting the "carry-over" of digits from one number to the next). Perhaps this is just a coincidence .....
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What is the relation between those number other than this property? I am not able to see any pattern.
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I'm beginning to think that there really isn't a pattern. There isn't a 3-digit perfect square of the form $\overline{abc}$ where $a$ and $\overline{bc}$ are (non-zero) perfect squares, (let alone $\overline{bc} = 16$ ). Also, there is no 7-digit perfect square with $625$ as the first three digits. So the "curious" sequence $1681,81225,225625$ is probably just a coincidence.
(I tried OEIS as well and didn't get any matches.)
Great solution! I love those Pythaogrean triples of the form $n, \frac{n^2 - 1}{2}, \frac{n^2 + 1}{2}.$
The question should have been more clarified. I THOUGHT FIRST 2 DIGITS AND LAST 2 DIGITS ARE INDIVIDUALLY PERIOD SQUARES. So, I typed 9409 which is the square of 97. First two digits, 9 and 4 are perfect squares of 3 and 2 respectively. Last two digits and 9 are perfect squares of 0 and 3, respectively. SO, THE QUEST I ON IS WRONG!!
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I am sorry for the confusion, and reformulated the question.
And 9409 is an interesting number because of all these squares!
(In your interpretation of the problem, though, 9409 is still not correct because the last two digits were to be a square not equal to zero...)
You are right, Ashish. There is a gross error in the wording or formulation of the question itself. In fact many of the questions posted on this website have serious flaws and yet somebody comes up with an answer that they can't properly explain. Let's see what is really wrong with this question:
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Hey, big fonter-- I think this was clarified above. Also, most people seem to know perfectly well what is meant. Language is only as precise as necessary, in order to be as concise as is efficient. :)
Why did you assumed that the number is in the form $z^2 = 100x^2 + y^2$ ? It's because the first and last two digits are perfect squares? I thought this way: $z = 10x + y$ $z^2 = 100x^2 + 20xy + y^2$ And my other doubt is why you did $10x + z = 20x$ .
Thank you!
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I wrote $z^2 = 100x^2 + y^2$ because the first and last pairs of digits must be squares. Writing $z = 10x + y$ is simply not very helpful, because it generates the mixed product $20xy$ , which affects the middle two digits.
Also, I did not assume that $10x + z = 20x$ ; rather, $10x + z > 20x$ because $z > 10x$ . I knew that because $z^2 = 100x^2 + y^2$ so that $z^2 > 100x^2$ .
Let the number be abcd.
abcd = 1000a +100 b + 10 c + d
A/q, abcd = z^2 ( say )
ab = x^2 = 10 a + b
or, 100 x^2 = 1000 a +100 b
And, cd = y^2 = 10 c + d
Thus, abcd = 100 x^2 + y^2 = z^2
There is a gross error in the wording or formulation of the question itself. In fact many of the questions posted on this website have serious flaws and yet somebody comes up with an answer that they can't properly explain. Let's see what is really wrong with this question:
Let the number be abcd, where ab=n^2 and cd=m^2, where m and n are not less than 4 because their squares are two digit numbers. So, we have,
10a+b=n^2, and 10c+d=m^2, and the number abcd=N=1000a+100b+10c+d,
N=100(10a+b)+(10c+d)=100n^2+m^2,
N is also a perfect square, so it will be of the form (10n+p)^2=100n^2+20np+p^2.
Equating the two gives,
m^2 = p^2 + 20np.
Since m^2 can't be more than 81, the numbers n and p can not be more than 4, but as discussed above, n can not be less than 4 also. So n = 4.
Now we have,
m^2 = p^2 + 80p >= 81, but m^2 <= 81,
So, m^2 = 81, m = 9.
Answer, N = 100n^2 + m^2 = 1681.
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Let the number be of the form $z^2 = 100 x^2 + y^2,$ with $1 \leq x,y \leq 9$ . The fact that the numbers may not start in zero requires in fact $x \geq 4$ .
Now the difference between two squares $a^2 - b^2 = (a-b)(a+b) \geq a+b$ . Therefore, $y^2 = z^2 - 100x^2 = z^2 - (10x)^2 \geq 10x+z \geq 20x.$ But we know that $y \leq 9\ \therefore\ y^2 \leq 81\ \therefore\ 20x\leq 81\ \therefore\ x\leq 4.$ But we also know that $x \geq 4$ . It follows that $x = 4$ .
Thus $z^2 = 1600 + y^2$ . Now $41^2 = 1681$ is the first square greater than 1600, and there are no other squares starting in 16xx. Thus the only solution is $\boxed{1681}$ , corresponding to the Pythagorean triple $41^2 = 40^2 + 9^2.$ (This problem is inspired by the increase of my followers from 40 to 41.)