Including 2500, it's 42,785.
5
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 sum of their squares is 10.
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.
Yes. 6 squared+71 squared equals 5077
Yes, 5041 and 36.
no
8081 can be the sum of two perfect squares because its perfect squares are 41 x41+80x80=1681+6400. Answer=1681+6400
No. The closest integers which can are 4001 (= 40² + 49²) and 4005 (= 6² + 63²).
The only squares of perfect squares in that range are 1, 16, and 81.
64 and 36.
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)
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.
Including 2500, it's 42,785.
9+16+25= 50
5