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Three Squares a Day
Find three numbers, all under 500, where the sum of all three is a perfect square and the sums of any two of them are also perfect squares.
Answer
41, 80, 320Hide Answer Show Answer
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Comments
Neat! Good Job!
Another solution: 88, 168, 273. 88 + 168 = 256, 88 + 273 = 361, 168 + 273 = 441, 88 + 168 + 273 = 529.
So easy!
lol! HARD!
GREAT job!!
lol! HARD!
GREAT job!!
I figured since it didn't say the numbers had to be different, there should be a solution in the form A, A, B where A + A is a square number. That means that any integer solutions for n in the equation:
n^2 = (n - x)^2 + (n - y) - (n - x)^2/2
gives
A = (n - x)^2/2
B = (n - y)^2 - A
The first equation expands to:
n^2 = n^2 - 2nx + x^2 + n^2 - 2ny + y^2 - (n^2 - 2nx + x^2)/2
multiply by 2:
2n^2 = 2n^2 - 4nx + 2x^2 + 2n^2 - 4ny + 2y^2 - n^2 + 2nx - x^2
combine terms:
2n^2 = 3n^2 - 2nx + x^2 - 4ny + 2y^2
subtract n^2 from each side:
0 = n^2 - 2nx + x^2 - 4ny + 2y^2
I was pretty sure if there was a solution, there would be a solution where x = 1, so I plugged in 1 for x and tried 2 for y:
0 = n^2 - 2n + 1 - 8n + 8
0 = n^2 - 10n + 9
Using the quadratic equation to solve for n:
n = (10 +/- SQRT(100 - 36))/2
n = (10 +/- /2
n = 1 and n = 9
Plugging 1 and 9 back in to calculate A and B:
A = (1 - 1)^2/2 = 0
B = (1 - 2)^2 - A = 1
A = (9 - 1)^2/2 = 32
B = (9 - 2)^2 - A = 17
Give two solutions: 0, 0, 1 and 32, 32, 17. Check:
0 + 0 = 0
0 + 1 = 1
0 + 0 + 1 = 1
32 + 32 = 64
32 + 17 = 49
32 + 32 + 17 = 81
This also means that where z is an integer, any set of numbers in the form:
A = (0z)^2/2 = 0
B = (1z)^2 - A = z^2
or
A = (8z)^2/2 = 32z^2
B = (7z)^2 - A = 49z^2 - A
are solutions.
Plugging various values for z into both equations gives:
0: 0, 0, 0 and 0, 0, 0
1: 0, 0, 1 and 32, 32, 17
2: 0, 0, 4 and 128, 128, 68
3: 0, 0, 9 and 288, 288, 153
4: 0, 0, 16 and 512, 512, 272
. . .
I went ahead and plugged the formulas into Excel to look for other solutions for x and y. Other than the multiples produced by the above equations, I also found integer solutions for (x, y) of (1, 16), (1, 9, and (1, 576) each of which also produces solutions that are a multiples.
I plugged 2,16,98,576 into OEIS and found this is sequence A082639 (http://www.research.att.com/~njas/sequences/ A082639) and that the next three values for y are 3362, 19600, and 114242.
n^2 = (n - x)^2 + (n - y) - (n - x)^2/2
gives
A = (n - x)^2/2
B = (n - y)^2 - A
The first equation expands to:
n^2 = n^2 - 2nx + x^2 + n^2 - 2ny + y^2 - (n^2 - 2nx + x^2)/2
multiply by 2:
2n^2 = 2n^2 - 4nx + 2x^2 + 2n^2 - 4ny + 2y^2 - n^2 + 2nx - x^2
combine terms:
2n^2 = 3n^2 - 2nx + x^2 - 4ny + 2y^2
subtract n^2 from each side:
0 = n^2 - 2nx + x^2 - 4ny + 2y^2
I was pretty sure if there was a solution, there would be a solution where x = 1, so I plugged in 1 for x and tried 2 for y:
0 = n^2 - 2n + 1 - 8n + 8
0 = n^2 - 10n + 9
Using the quadratic equation to solve for n:
n = (10 +/- SQRT(100 - 36))/2
n = (10 +/- /2
n = 1 and n = 9
Plugging 1 and 9 back in to calculate A and B:
A = (1 - 1)^2/2 = 0
B = (1 - 2)^2 - A = 1
A = (9 - 1)^2/2 = 32
B = (9 - 2)^2 - A = 17
Give two solutions: 0, 0, 1 and 32, 32, 17. Check:
0 + 0 = 0
0 + 1 = 1
0 + 0 + 1 = 1
32 + 32 = 64
32 + 17 = 49
32 + 32 + 17 = 81
This also means that where z is an integer, any set of numbers in the form:
A = (0z)^2/2 = 0
B = (1z)^2 - A = z^2
or
A = (8z)^2/2 = 32z^2
B = (7z)^2 - A = 49z^2 - A
are solutions.
Plugging various values for z into both equations gives:
0: 0, 0, 0 and 0, 0, 0
1: 0, 0, 1 and 32, 32, 17
2: 0, 0, 4 and 128, 128, 68
3: 0, 0, 9 and 288, 288, 153
4: 0, 0, 16 and 512, 512, 272
. . .
I went ahead and plugged the formulas into Excel to look for other solutions for x and y. Other than the multiples produced by the above equations, I also found integer solutions for (x, y) of (1, 16), (1, 9, and (1, 576) each of which also produces solutions that are a multiples.
I plugged 2,16,98,576 into OEIS and found this is sequence A082639 (http://www.research.att.com/~njas/sequences/ A082639) and that the next three values for y are 3362, 19600, and 114242.
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