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I would like to know if it has been proved that :

  • There are no $a$, $b$ and $c$, all prime numbers, such that $a^2 + b^2 = c^2$
  • There are no $a$, $b$, $c$ and $d$, all prime numbers, such that $a^2 + b^2 + c^2 = d^2$
  • There are no $a$, $b$, $c$, $d$ and $e$, all prime numbers, such that $a^2 + b^2 + c^2 + d^2 = e^2$

Thanks for your answers.

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  • $\begingroup$ For the first, we must have $a=2$ or $b=2$. $\endgroup$
    – lhf
    Aug 25, 2014 at 12:15
  • $\begingroup$ But there is no Pythagorean triplets (a, b, c) with $a = 2$ or $b = 2$. $\endgroup$ Aug 25, 2014 at 12:16
  • $\begingroup$ Yes, thats why the first one is trivial. $\endgroup$
    – user153012
    Aug 25, 2014 at 12:17
  • $\begingroup$ But why we must have $a = 2$ or $b = 2$ ? $\endgroup$ Aug 25, 2014 at 12:18
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    $\begingroup$ For the second: If $a$ is not divisible by $3$, then $a^2=1\pmod{3}$. For the third: If $a,b,c,d$ are all odd, then $d$ is even. $\endgroup$ Aug 25, 2014 at 12:24

1 Answer 1

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It turns out that the (full) solution is longer than a comment. So here it goes.


Suppose that for $a,b,c,d$ primes, $$a^2+b^2+c^2=d^2$$ Obviously $d=3$ is not a solution. That implies $$a^2+b^2+c^2\equiv 1\pmod{3}.$$ So exact two of $a,b,c$ are equal to $3$. We are left to solve $$d^2-a^2=(d-a)(d+a)=18.$$ But $d-a$ and $d+a$ are of the same parity and $18$ cannot be factorized into such a product.


For the third equations, note that $n^2\equiv 1\pmod{4}$ if $n$ is odd. Thus if $$a^2+b^2+c^2+d^2=e^2,$$ then exactly $3$ of $a,b,c,d$ are $2$. We are left to solve $$(e-d)(e+d)=12=6\cdot 2.$$ It follows that $e+d=6$, and $e-d=2$. That means $e=4$, not a prime.

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  • $\begingroup$ Yes, you are right. It is nearly the same proof I found on my own. $\endgroup$ Aug 25, 2014 at 12:57

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