Puzzle time
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wrote on 15 Nov 2020, 01:14 last edited by Klaus
Find all natural numbers x and y such that the sum of the square roots of x and y is the square root of 2009.
No computer is needed.
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wrote on 15 Nov 2020, 01:50 last edited by
@Klaus said in Puzzle time:
Find all natural numbers x and y such that the sum of the square roots of x and y is 2009.
All the square roots or just the positive ones?
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wrote on 15 Nov 2020, 03:12 last edited by jon-nyc
||Assuming just the positive roots then it’s just (1^2, 2008^2), (2^2, 2007^2), (3^2, 2006^2), ....etc.||
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||Assuming just the positive roots then it’s just (1^2, 2008^2), (2^2, 2007^2), (3^2, 2006^2), ....etc.||
wrote on 15 Nov 2020, 03:47 last edited by||I don’t think your answer is correct. For example, the square root of one is one. And the square root of 2008 is almost 45. Add those two together and only equal 46
Not sure of correct answer however. Lol46||
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wrote on 15 Nov 2020, 04:08 last edited by jon-nyc
||You misread it. I squared those numbers for ease of notation and understanding the sequence. The real numbers would be (1, 4,032,064), (4, 4,028,049), (9, 4,024,036)...||
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||You misread it. I squared those numbers for ease of notation and understanding the sequence. The real numbers would be (1, 4,032,064), (4, 4,028,049), (9, 4,024,036)...||
wrote on 15 Nov 2020, 04:12 last edited by@jon-nyc ah okay.
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wrote on 15 Nov 2020, 09:27 last edited by Klaus
Sorry, I screwed up. The sum of those numbers should be the square root of 2009. I fixed it above. The previous version is trivial, as Jon pointed out.
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@Klaus said in Puzzle time:
Find all natural numbers x and y such that the sum of the square roots of x and y is 2009.
All the square roots or just the positive ones?
wrote on 15 Nov 2020, 09:51 last edited by@jon-nyc said in Puzzle time:
@Klaus said in Puzzle time:
Find all natural numbers x and y such that the sum of the square roots of x and y is 2009.
All the square roots or just the positive ones?
Just the positive ones.
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wrote on 15 Nov 2020, 11:54 last edited by
||Since the positive square root of 2009 is not a natural number, there is zero natural number for x and y such that they sum to the positive square root of 2009.||
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wrote on 15 Nov 2020, 12:03 last edited by
Ax that’s not the problem. I’ll restate it:
Solve for sqrt(x) + sqrt(y) = sqrt(2009) where x,y are natural numbers
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wrote on 15 Nov 2020, 13:06 last edited by
|| (41 * a^2, 41 * b^2) where a,b are natural numbers with a+b=7.
Or to spell it out, (41,1476), (164,1025), (369,656) and then the three number pairs that have those numbers reversed||
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wrote on 15 Nov 2020, 14:47 last edited by
That looks good, Jon. Can you show how you got that solution?
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wrote on 15 Nov 2020, 16:55 last edited by jon-nyc
2009^1/2 simplifies to 7*sqrt(41)
So the two addends need to be in the form a * sqrt(41) and b * sqrt(41) with a and b summing to 7.
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wrote on 15 Nov 2020, 19:14 last edited by