Is there a sum or perfect square of 2013 positive and perfect squares that are different from each o

Why did you ask again?

This is right, the idea is that any even number can be the right-angled side of a right-angled triangle, and the other right-angled edge and hypotenuse are even. This number is called the Pythagorean number, and we start with (3,4,5).

a=m -n, b=2mn, c=m +n a, b is the right-angled edge, and c is the hypotenuse. m, n are integers.

I got 5, but 5 is an odd number, so I multiply all the sides by 2 and get it.

Then set 10 2mn, and find that we can get m=5, n=1a=m -n, b=2mn, c=m +n.

26 is an even number, let 26 2*13*1, find that you can make m=13, n=1a=m -n, b=2mn, c=m +n find that 170 is an even number, so 170=2*m*n, find that you can make m=85, n=1a=m -n, b=2mn, c=m +n find that 7226 is an even number, so 7226=2*m*n, find that you can make m=3613, n=1

a=m -n, b=2mn, c=m +n, and so on, if an odd number is found in the middle, then the length of all sides of the above process is 2, and the problem can be solved.

It's like I dealt with 5.

Not good, so change to.

So there is.

Let the integers a,b,c satisfy a 2+b 2 = c 2, then if n is an integer and n>1, there must be (an) 2+(bn) 2 = (cn) 2

So there are infinite groups that meet the different conditions.

a^2+b^2=c^2, a^2-c^2=d^2

The formula for the integer root of the Pythagorean equation is obtained from a 2 + b 2 = c 2: a, b one of them is an odd number, and the other is an even number, you may wish to make a odd number and b an even number, so that c is an odd number.

a=m 2-n 2, b=2mn, c=m 2+n 2, where (m,n)=1, so c-a=2n 2

In the same way, by c 2 + d 2 = a 2, because c is an odd number and a is an odd number, it is b even, and we get:

c=p 2-q 2, d=2pq, a=p 2+q 2, so a-c=2q 2

The addition of the two formulas has: 2n 2+2q 2=0, and can only have: n=q=0, so b=d=0 a=c

Therefore there is no non-zero solution.

I'll give it a try.

Proof: For the odd perfect square number 2k+1, we construct 2k+1=m-n =(m+n)(m-n)m-n=1

Therefore m=k+1 , n=k

For even numbers exactly squared 4k+4, we construct.

4k+4=m²-n²=(m+n)(m-n)m-n=2

Therefore m=k+2 , n=k

Thus, for any perfect square number k, we can construct the appropriate n such that k + n = m

Let the first perfect square number k1 construct k2 so that k1 +k2 = m2 construct k3 and make m2 +k3 = m3 , construct k4....And so on until K2011

Then there is k1 +k2 +k3 +k2011²=m2011²。Certification.

With the construction method, you only need to find a set of matching ones.

It is also possible to use the inductive method to prove that it holds true on a smaller scale and then expand it.

exists.

For example (10 223-5).

Of these, there are 222 9s and 222 0s).

The sum of the numbers is.

P.S. The meaning of a b is the power b of a.