Including 2500, it's 42,785.
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.
1x1=12x2=43x3=94x4=165x5=256x6=367x7=498x8=649x9=8110x10=10011x11=12112x12=144So to sum it all up, the first 12 perfect squares are1,4,9,16,25,36,49,64,81,100,121,144.
5
2500
The sum of their squares is 10.
The only squares of perfect squares in that range are 1, 16, and 81.
8081 can be the sum of two perfect squares because its perfect squares are 41 x41+80x80=1681+6400. Answer=1681+6400
The number 1 is a perfect square that is equal to the sum of the perfect squares that precede it, as there are no perfect squares before it (0 is not considered a perfect square in this context). Additionally, the number 5 is another perfect square, specifically (2^2), which equals the sum of the perfect squares 0 (which is (0^2)) and 1 (which is (1^2)). However, the most straightforward example is 1.
9+16+25= 50
no
64 and 36.
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
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.
1x1=12x2=43x3=94x4=165x5=256x6=367x7=498x8=649x9=8110x10=10011x11=12112x12=144So to sum it all up, the first 12 perfect squares are1,4,9,16,25,36,49,64,81,100,121,144.
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.
5
2500