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).
64 and 36.
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.
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