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if postive integer $a,b,c,d$ such $$a^{14}+b^{14}+c^{14}=d^{14}$$

show that $$7\mid abcd$$

My idea: maybe this problem can use Fermat's little theorem. but I can't prove it.Thank you

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One can verify that $n^{14}$ is always congruent to one of $0,1,18,30$ modulo $49$, and is congruent to $0\pmod{49}$ exactly when $7\mid n$. One can then further verify that the only solutions to $w+x+y\equiv z\pmod{49}$ with $w,x,y,z\in\{0,1,18,30\}$ are: when two of $w,x,y$ equal $0$ and the other one equals $z$; and when $\{w,x,y\}=\{1,18,30\}$ and $z=0$.

(Why look modulo $49$? Well, we care about divisibility by $7$, to start with; and we know that $(\Bbb Z/49\Bbb Z)^\times$ is a cyclic group of order $42$, which is a multiple of $14$, and hence the $14$th powers modulo $49$ will be very few.)

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$$a^{14}-a^2≡0 \pmod7$$ $$a^{14}+b^{14}+c^{14}-(a^2+b^2+c^2 )=d^{14}-d^2+[d^2-(a^2+b^2+c^2 ) ]$$ Hence $ d^2-(a^2+b^2+c^2 )≡0 \pmod7$. By Fermat’s little theorem $$(a^2+b^2+c^2 )-d^2=7k+m+n+h-s , \quad m,n,h,s=1,4,9$$ If none of $a,b,c,d$ is divisible by $7$, it is an easy task to check that then $m+n+s-h $ is not divisible by $7$.Therefore, $7|abcd$.

P.Ranawaka

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