Question 1185809: What is the fourth positive integer N such that the product of any two increased by 1 of the numbers 3,8,21,N is a square
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Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20056)   (Show Source):
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The products of two numbers among 3,8,21 are 24,63,168.
None of those products increased by 3,8, or 21 are squares.
So those products 24,63,168 must depend upon adding N to them to
give a square sum. Therefore, if there is any solution to the problem:
24+N = a2 for some positive integer a.
63+N = b2 for some positive integer b.
168+N = c2 for some positive integer c.
Thus N = a2-24 = b2-63 = c2-168
That only has one solution, a=5, b=8, and c=13, which makes N=1
However, that means the product 21N is 21. But, no matter which of these:
3,8,21,1 we add to the product 21 gives a square for a sum.
Thus there is no possible positive integer N meeting the conditions
that the product of any two increased by 1 of the numbers 3,8,21,N is a square.
This proves there is no solution. Could there be a typo, perhaps in the number
21?
Edwin
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
I am guessing that the other tutor misinterpreted the problem because of the awkward way it was stated.
Your statement:
What is the fourth positive integer N such that the product of any two increased by 1 of the numbers 3,8,21,N is a square
I believe, for clarification, that the intended statement of the problems should be this:
What is the fourth positive integer N such that the product of any two of the numbers 3,8,21,N, increased by 1, is a square
If that is the intended meaning of the problem, then the given conditions are satisfied:
3*8+1=25 = 5^2
3*21+1=64 = 8^2
8*21+1=169 = 13^2
If that is what the problem was intended to say, then re-post....
I note that the numbers involved in the problem are 3, 5, 8, 13, and 21, which are successive numbers in the Fibonacci sequence; so I suspect the answer might be found in the Fibonacci sequence. But I haven't found it.
55 ALMOST works:
21*55+1=1156=34^2
8*55+1=441 = 21^2
but 3*55+1=166 is not a perfect square....
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