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Question 1151144: Five integers between 440 and 500 can each be expressed as the sum of two
different squares in two different ways. Find their median.
Found 2 solutions by jim_thompson5910, Edwin McCravy:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
The 5 numbers that meet the description are 442, 445, 481, 485, 493
Based on what is shown below:
1^2 + 21^2 = 1 + 441 = 442
9^2 + 19^2 = 81 + 361 = 442
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2^2 + 21^2 = 4 + 441 = 445
11^2 + 18^2 = 121 + 324 = 445
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9^2 + 20^2 = 81 + 400 = 481
15^2 + 16^2 = 225 + 256 = 481
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1^2 + 22^2 = 1 + 484 = 485
14^2 + 17^2 = 196 + 289 = 485
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3^2 + 22^2 = 9 + 484 = 493
13^2 + 18^2 = 169 + 324 = 493
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I used a perl script to search for all the integers between 440 and 500, excluding the endpoints (see note below), such that the integer could be broken down into a sum of two distinct squares.
Note: we can express 500 as a sum of two different squares in two different ways
4^2 + 22^2 = 16 + 484 = 500
10^2 + 20^2 = 100 + 400 = 500
but that would mean we're dealing with 6 values instead of 5. So it appears we are ignoring the endpoints 440 and 500.
We are also ignoring 450
3^2 + 21^2 = 9+441 = 450
15^2 + 15^2 = 225+225 = 450
because the second summation line has us adding the same square to itself, ie we dont have two distinct square numbers here.
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Once again the five numbers are: 442, 445, 481, 485, 493
The median is the middle most value. In this case, it would be the value in the third slot.
Answer: 481
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
That's false because, as you can see below, SIX, not five, integers between 440
and 500 can be written as the sum of two squares in two different ways. If we
include 500, then SEVEN integers can be.
1² + 21² = 442
9² + 19² = 442
2² + 21² = 445
11² + 18² = 445
3² + 21² = 450
15² + 15² = 450
9² + 20² = 481
15² + 16² = 481
1² + 22² = 485
14² + 17² = 485
3² + 22² = 493
13² + 18² = 493
4² + 22² = 500
10² + 20² = 500
Edwin
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