Puzzle time

[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]