Saurav B.
asked • 08/22/14number of positive integers n for which n^2+96 is a perfect square is??
number of positive integers n for which n^2+96 is a perfect square is??
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Phillip R. answered • 08/23/14
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492 - 482 = 2401 - 2304 = 97
Therefore any possible answer must be less than 49.
The positive integers n for which n2 + 96 = y2 where y is an integer are
2, 5, 10, and 23
2 and 10
5 and 11
10 and 14
23 and 25
a somewhat more formal apprPacheco is to say that (n+1)^2 - n^2 = 96
(n+2)^2 - n^2 = 96
.
.
.
(n+9)^2 - n^2 = 96
there is no need to go beyond n+9 because the only possible solution is n<1
so we can solve for n in those cases were 2i + i^2 is even
This occurs for n= 2, 5, 10, and 23
it looks like this
4n + 4 = 96
8n + 16 = 96
12n + 36 = 96
16n + 64 = 96
Dattaprabhakar G.
This comment is for Phillip R. Thank you in advance.
The way I was approaching this interesting problem is as follows:
Set n = 2+m. We wish to find values for whole numbers m such that (m+2)2 + 96 = y2.
That is, m2 + 4m + 100 = y2. We want to find values for m such that m2 + 4m + 100 is a perfect square. (of y)
You and I know that m= 0, 3, 8 and 21 yield perfect squares. However, there may be some m > 21 that also yields a perfect square. The question is how to go about finding such m's. Is there an algorithm?
The canonical representation of a perfect square does not seem to help here. Please advise.
BTW, I did not understand the word apprPacheco in "a somewhat more formal apprPacheco is to say that (n+1)^2 - n^2 = 96." Please help.
Once again, thank you, Phillip.
Dattaprabhakar (Dr. G.)
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08/23/14
Phillip R.
The word should be approach. We know there are no values greater than 23 because 23 is the solution to the equation 4n + 4 = 96 and there is no positive integer solution to the equation 2n + 1 = 96. Because (n +1)^2 - n^2 = 2n + 1, it is the largest possible
n that could satisfy the criteria n^2 + 96 = y^2
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08/23/14
Dattaprabhakar G.
Thank you, Phillip. Dr, G.
Report
08/23/14
Dattaprabhakar G. answered • 08/23/14
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Saurav:
You want n2 + 96 to be a perfect square. Obviously, this perfect square must be a number larger than 96, The first perfect square that comes to mind is 100 = (10)2.Take n = 2. That is one possible value of n ( a positive integer) for which n2 + 96 is a perfect square. Is it the ONLY value? I will have to think on that one.
Dattaprabhakar (Dr, G,)
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Dattaprabhakar G.
Let Q be a given positive integer.
Result 1:A necessary and sufficient condition that a positive integer n exists such that n2 + Q = y2, for some positive integer y, is that for some positive integer m, (Q – m2 ) /2m is a positive integer.
Now if (Q – m2 ) / 2m is a positive integer we can set n = (Q – m2 ) / 2m. Then, n2 + Q = y2 = (n + m)2. Conversely, if n and m exist such that n2 + Q = (n + m)2 is satisfied, then it follows that (Q – m2 ) / 2m is a positive integer (= n).
Proof: Straightforward verification.
Proof: Write (Q – m2 ) / 2m = (1/2)(Q/m – m). Then (1/2)(Q/m - m) – (1/2)[Q/(m+1) - (m+1) ] = (1/2)[Q/m(m+1) + 1] > 0.
Remark: Consequence of Result 2 is that there is only a limited number of values, if at all, for which n2 + Q = y2 holds.
Some more examples:
1. Q = 45. Then for m = 1, (Q – m2 ) / 2m = 22. Hence 222 + 45 = 232.
2. Q = 8. It can be easily verified that positive integers n and y such that n2 + Q = y2, do not exist!
3. Q = 38. It can be easily verified that positive integers n and y such that n2 + Q = y2, do not exist!
5. Q = 24 m = 1,DOES NOT WORK, m = 2, (Q – m2 ) / 2m= 20/4 = 5. Hence 52 + 24 = 72.
08/24/14