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For primes $ p \geq 17 $, prove that there exist $ a \in \mathbb{Z}_P^*$ such that $\left( \frac{a}{p} \right) = -1$ and $ord_p(a) < p-1$.

$\mathbb{Z}_P^* = \{1, 2, ..., p-1\}$

Please give me some suggestions.

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    I suggest you try $a=1$, $a=2$, etc., until you find one that works. – 2017-01-29
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    @GerryMyerson This way, we cannot show the existence of such an $a$ for EVERY prime $p\ge 17$ – 2017-01-29

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The claim is not true for $p=17$ (or any other Fermat-prime)

All the quadratic non-residues modulo $17$, which are $3,5,6,7,10,11,12,14$ , have order $16$ modulo $17$

Suppose, $\phi(p-1)<\frac{p-1}{2}$, (which is true for any odd prime not being a fermat prime, as I will show later)

It is well known that there are $\phi(p-1)$ primitive roots modulo $p$ (Order modulo $p$ is $p-1$), but there are $\frac{p-1}{2}$ quadratic non-residues modulo $p$. Hence, there are more qudratic-non-residues than primitive-roots, hence there must be at least one quadratic non-residue which is not a primitive root

So, what remains to show is the inequality $\phi(p-1)<\frac{p-1}{2}$ , but this is not difficult because the even numbers cannot be coprime to $p-1$, if $p$ is odd, so $\phi(p-1)\le \frac{p-1}{2}$ is clear.

And it is well known that $\phi(p-1)=\frac{p-1}{2}$ is true only for Fermat-primes $p$.

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    In case that $\phi(p−1) = \frac{p-1}{2}$, number of quadratic-non-residues is equal to number of primitive roots, it's not necessary that there will be $a$ satisfying the problem conditions, am I right ? – 2017-01-29
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    Yes, and at least the primes $p=17,257,65537$ are counter-examples. If you replace $17$ by $65539$, a counterexample (if existent) must have more than $2\cdot 10^9$ digits, so the claim is true for every odd prime with reasonable magnitude. – 2017-01-29
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    Since it's not true for $p=17$ and other Fermat-primes, can we conclude that the problem statement is not true ? – 2017-01-30
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    Yes, it is (probably) true for $p\ge 65539$, but false for $p=17,257,65537$. So in the stated form, it is false. – 2017-01-30
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    Thank you very much for your thorough explanation, Peter. – 2017-01-30