0
$\begingroup$

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?

Improve this question
$\endgroup$
2
$\begingroup$

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.

Improve this answer
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $A^2+B^2=C^2$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.