prove there exist postive integers $a,b$ such $p^2|a^2+ab+b^2$ Problem 1: Let prime $p\equiv 1\pmod 3$.show that:there exist postive integers $a\le b<p$ such
$$p^2|a^2+ab+b^2$$
I have only prove there $a,b$ such $$p|a^2+ab+b^2$$
Problem 1 from this:
Problem 2.3 (Noam Elkies). Prove that there are infinitely many triples $(a,b,p)$ of integers, with $p$ prime and $0<a\leq b<p$, for which $p^5$ divides $(a+b)^p-a^p-b^p$.

The key claim is that if $p\equiv1\pmod3$, then $$p(x^2+xy+y^2)^2\;\mathrm{divides}\;(x+y)^p-x^p-y^p$$ as polynomials in $x$ and $y$. $\color{red}{\underline{\color{black}{\text{Since it's known that}}}}$ one can select $a$ and $b$ such that $\color{red}{\underline{\color{black}{p^2\mid a^2+ab+b^2}}}$, the conclusion follows. (The theory of quadratic forms tells us we can do it with $p^2=a^2+ab+b^2$; Thue's lemma lets us do it by solving $x^2+x+1\equiv0\pmod{p^2}$.)
To prove this, it is the same to show that $$(x^2+x+1)^2\;\mathrm{divides}\;F(x)\overset{\mathrm{def}}=(x+1)^p-x^p-1.$$ since the binomial coefficients $\binom pk$ are clearly divisible by $p$. Let $\zeta$ be a third root of unity. Then $F(\zeta)=(1+\zeta)^p-\zeta^p-1=-\zeta^2-\zeta-1=0$. Moreover, $F'(x)=p(x+1)^{p-1}-px^{p-1}$, so $F'(\zeta)=p-p=0$. Hence $\zeta$ is a double root of $F$ as needed.
(Incidentally, $p=2017$ works!)
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This relies on the following statement: if some residue class $a$ is a square mod $p$, then it is a square mod $p^2$. 
Indeed, if $b^2=a+tp [p^2]$, then $(b+kp)^2=a+(t+2k)p [p^2]$. 
So you know that there is some $x$ such that $p|x^2+x+1$, thus $-3=(2x-1)^2$ is a square mod $p$, thus mod $p^2$, and if $y^2=-3[p^2]$, then $a=(1+y)/2$ (mod $p^2$) and $1$ work.
