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Determine if there are 2041 distinct natural numbers such that the sum of their squares is a perfect square.

Can anyone please help me to solve this problem?

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Hint Any multiple of $4$ is the difference of two even perfect squares.

Hint 2 You can make the sum of $2040$ perfect squares a multiple of $4$. You can moreover chose the numbers in such a way that they don't repeat and the two numbers in the above hint are not in this set.

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  • $\begingroup$ Can you explain it through the numbers and equations because I didn't get how to use these hints thanks $\endgroup$ – Nariman Zendehrooh Nov 12 '17 at 1:06
  • $\begingroup$ Can you please respond $\endgroup$ – Nariman Zendehrooh Nov 13 '17 at 4:09
  • $\begingroup$ @NarimanZendehrooh I gave you all the information you need to solve the problem. Pick first 2040 numbers such that the sum of their squares is a multiple of 4. Then make their sum a difference of two squares. $\endgroup$ – N. S. Nov 13 '17 at 4:55
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This is a Pythagorean Theorem problem. There are two general formulas that may be used for generating Pythagorean triples.

Euclid's formula: $ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$ which generates trivial triples such as $(2,0,2)$, primitives such as $(3,4,5)$, and square multiples of primitives such as $(12,16,20)$ or $(27,36,45)$.

Another formula (mine) is \begin{equation} A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2 \end{equation}

which generates only the subset of triples where $GCD(A,B,C)$ is an odd square as shown below. \begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline \end{array}

By inspection, we can see that there are infinite combinations of $(n,k)$ that generate distinct valid Pythagorean triples which, by definition, have the property of $A^2+B^2=C^2$.

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