4
$\begingroup$

Find all $p_1,\dots,p_8$ so that $$ p_1^2+\dots+p_7^2 = p_8^2 $$ What I have done so far:
Since an odd number $n$ has $n^2 \equiv 1$ (mod $4$) then the only two possibilities are if $2$ or $6$ of the primes on the left are $2$. When $6$ of them are $2$, then I found a solution when $p_7 = 5$ and $p_8=7$. I cannot figure out the other case though. I got to $$ p_3^2 + p_4^2 + p_5^2 + p_6^2 + p_7^2 = p_8^2-8 $$ but that's it. But I cannot find a way to find them all or prove none exist.

$\endgroup$
  • $\begingroup$ Your equation for the second case is not correct. If 2 of them are 2, the right hand side is $p_8^2 - 8$. So your last example is also not correct. $\endgroup$ – user113102 May 28 at 21:08
  • $\begingroup$ You're right. My bad $\endgroup$ – Kristin Petersel May 28 at 21:15
  • 1
    $\begingroup$ Try working mod 8 in the second case. $\endgroup$ – user113102 May 28 at 21:36
4
$\begingroup$

As user113102 hints, if $n$ is odd, then $n^2 \equiv 1 \bmod 8$. Thus, in the equation $$p_3^2 + p_4^2 + p_5^2 + p_6^2 + p_7^2 = p_8^2 - 8$$ where all $p_i$ are odd primes, the LHS is congruent to $5$ mod $8$ and the RHS is congruent to $1$, so there is no solution.

It's easy to see, therefore, that the solution $p_1 = \cdots = p_6 = 2, p_7 = 5, p_8 = 7$ is therefore the only solution to the general case, as larger values of $p_7$ and $p_8$ would give $p_8^2 - p_7^2 > 24$.

$\endgroup$

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.