Q: Can 2819 be expressed by the sum of 2 perfect squares?

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Yes. 6 squared+71 squared equals 5077

8081 can be the sum of two perfect squares because its perfect squares are 41 x41+80x80=1681+6400. Answer=1681+6400

The only squares of perfect squares in that range are 1, 16, and 81.

Since the last digit of 5077 is 7, the last digit of the perfect square numbers must be 1 and 6. So that we have 5077 = 712 + 62.

64 and 36.

No. The closest integers which can are 4001 (= 40² + 49²) and 4005 (= 6² + 63²).

It is Fermat's theorem on the sum of two squares. An odd prime p can be expressed as a sum of two different squares if and only if p = 1 mod(4)

5

Including 2500, it's 42,785.

9+16+25= 50

Yes, 5041 and 36.

It is not clear what the question means: there are 31 2-digit numbers that can be expressed as a sum of two squares.

The proposition in the question is simply not true so there can be no answer!For example, if given the integer 6:there are no two perfect squares whose sum is 6,there are no two perfect squares whose difference is 6,there are no two perfect squares whose product is 6,there are no two perfect squares whose quotient is 6.

The sum of their squares is 10.

the smallest is 1 and the next is 4, so add those together.

Only 25 which is, 52 = 32 + 42 (25 = 9 + 16)

If the regression sum of squares is the explained sum of squares. That is, the sum of squares generated by the regression line. Then you would want the regression sum of squares to be as big as possible since, then the regression line would explain the dispersion of the data well. Alternatively, use the R^2 ratio, which is the ratio of the explained sum of squares to the total sum of squares. (which ranges from 0 to 1) and hence a large number (0.9) would be preferred to (0.2).

25 = 9 + 16 There are many more sets like these. This one has the smallest numbers.

There is a calculation error.

No.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theoremNo.First of all, you can't write negative numbers as sums of perfect squares at all - since all perfect squares are positive.Second, for natural numbers (1, 2, 3...) you may need up to 4 perfect squares: http://en.wikipedia.org/wiki/Lagrange's_four-square_theorem

12 + 22 + 32 + 42 + 52 = 1 + 4 + 9 + 16 + 25 = 55

Here is a procedure that would do the job nicely: -- Make a list of all the perfect squares between 5 and 30. (Hint: They are 9, 16, 25, 36, and 49.) -- Find the sum by writing the numbers in a column and adding up the column.

split 10 in two parts such that sum of their squares is 52. answer in full formula

Sum of squares? Product?

the highest sum of the numbers is 17 and the lowest is 1. The only perfect squares in that range are 1,4,9, and 16. That means the following numbers will work: 10,13.18,22,27,31,36,40,45,54,63,72,79,81,88, and 90; that is 16 numbers

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