Find prime numbers : $p_1,p_2,\cdots,p_8$ satisfying : $p_1^2+\cdots+p_7^2=p_8^2$.

Form test in class of my brother .


We already have an example with repeats. Assume there are no repeats; now look at the equation modulo 4. Quadratic residues: $0,1$.

If 2 does not appear on the left hand side, then everything is congruent to 1 there and so we obtain $7\equiv 3$ in total. This is impossible.

If 2 does, then we have $4+6\equiv 2$. This is also impossible.

Hence there are no solutions where the LHS comes out as being a perfect square of anything in this case!


Take $p_1=p_2=\cdots=p_6=2$. You don't have to search very far to find $p_7=5$, which gives LHS=$49=7^2$.

  • $\begingroup$ Nice,It exsts $p_{1}<p_{2}<\cdots<p_{8}$,such that? $\endgroup$ – math110 Sep 9 '13 at 13:55
  • $\begingroup$ when $p_i$ are different , we need an proof? $\endgroup$ – Hung Nguyen Sep 9 '13 at 14:51
  • $\begingroup$ @HungNguyen I've already given a quick proof using modular arithmetic that there is no such solution! (PS: Obviously +1!) $\endgroup$ – Sharkos Sep 9 '13 at 18:43

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