Two square scale meeting point, when one is moved odd

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rajeshmarndi
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There are two square scale. That is, it has marking where there are square i.e marking at 0,1,4,9,16,25 and so on. When one scale is moved, it slide over the other. Now if one scale is moved odd number i.e say 123 , that is, it's zero is placed at 123 over the other scale, now.

Now, can one know, if the squares on the two scales meet more than once?

Atleast twice because for every odd number there will always be two consecutive square whose gap is an odd number i.e 123 is gap between squares 61 and 62. So square 62 in one scale will meet at 61 square on the other scale that is moved. Thanks.
 
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I interpret your question as "for a given number n, how many pairs of squared integers are there that have a difference of n?" As an example, n=15 leads to (1,16) and (49,64).

This number is always finite, and for odd n you found the largest pair already. In general there can be others as well, it depends on the number n.
 
mfb said:
In general there can be others as well, it depends on the number n.
If we take n(odd) as very large, is there a way we can tell if it will have atleast two squared integers pairs.
 
The difference between k2 and (k+3)2 is 6k+9=3(2k+3), the difference between k and (k+5)2 is 10k+25=5(2k+5) and so on. If a number can be written as such a product then there are squares with this difference.
The second factor can be every odd number larger than the first one (as k>0 unless we count 0 as square number). Every factorization of n will work apart from its square root. The only odd numbers without additional solutions are primes and squares of primes. If you count 0 as square number, then primes are the only exception.
 
For large numbers we cannot say if they will have atleast two square integers pairs. Because we cannot know its factorization i.e we cannot say its primality.