About quadratic residue modular p(p is prime)

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The question is this.

Suppose that p is a prime, $p\ge7$. Show that $(\frac np)=(\frac {n+1}p)=1$ for at least one number n in the set {1,2,...,9}.

I think seperating into two cases when n=1 and when n=4 will help me prove it. But I can't think further...

Plz, HELP ME!!!!

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Assume that the statement is not true. We clearly have $\left(\frac{1}{p}\right) = \left(\frac{4}{p}\right) = \left(\frac{9}{p}\right) = 1$. By assumption we must then have $\left(\frac{2}{p}\right) = \left(\frac{5}{p}\right) = \left(\frac{10}{p}\right) = -1$. Can you derive a contradiction from this?