Puzzle time

[jon-nyc]

||Assuming just the positive roots then it’s just (1^2, 2008^2), (2^2, 2007^2), (3^2, 2006^2), ....etc.||

[taiwan_girl]

|@jon-nyc

||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||

[jon-nyc]

@taiwan_girl

||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)...||

[taiwan_girl]

@jon-nyc ah okay.

[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.

[Klaus]

@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.

[Axtremus]

||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.||

[jon-nyc]

Ax that’s not the problem. I’ll restate it:

Solve for sqrt(x) + sqrt(y) = sqrt(2009) where x,y are natural numbers

[jon-nyc]

|| (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||

[Klaus]

That looks good, Jon. Can you show how you got that solution?

[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.

[Klaus]