Diophantine equation

Algebra Level 2

Are there infinitely many distinct positive integer solutions (i.e a ≠ b ≠ c ≠ x ≠ y ≠ z a\neq b\neq c\neq x\neq y\neq z ) to this system of equations? a + b + c = x + y + z a+b+c=x+y+z a 2 + b 2 + c 2 = x 2 + y 2 + z 2 a^{2}+b^{2}+c^{2}=x^{2}+y^{2}+z^{2}

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Freddie Hand by Freddie Hand , 18, United Kingdom

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1 solution

Achal Jain
Feb 14, 2017

According to me easiest solution would be to equate a = x , b = y , c = z a=x,b=y,c=z . Voila, you have infinite solutions.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks for reminding me! I forgot to say that none of a,b,c,x,y and z are equal! The proof is then a lot harder. Here's a possible solution- a=6, b=5, c=1, x=7 ,y=3, and z=2. You can induct all other solutions.

Freddie Hand - 4 years, 3 months ago

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