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!)
Image that replaced the text
 A: 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.
A: Alternate solution via Thue's Lemma

Proposition. Let $p\equiv 1\pmod 3$ be a prime. Then there is a solution to
  $$
x^2+x+1\equiv 0 \pmod{p^2}
$$

Proof. By Quadratic Reciprocity, $-3$ is a quadratic residue $\pmod p$, hence
$$
u^2 \equiv -3 \pmod p
$$
for some integer $u$. Setting $u\equiv 2c+1\pmod p$, this becomes
$$
c^2+c+1 \equiv 0 \pmod p
$$
So $c$ is a solution to $f(x):=x^2+x+1\equiv 0\pmod p$. Hensel's Lemma tells us that if
$$
f'(c)\not\equiv 0 \pmod p
$$
Then there exists a unique $d=c+kp$ such that
$$
f(d) \equiv 0 \pmod{p^2}
$$
Since
$$
f'(c)^2 = (2c+1)^2 \equiv -3 \pmod p
$$
If $f'(c)\equiv 0 \pmod p$ then $0\equiv -3\pmod p$ which implies $p=3$. This contradicts $p\equiv 1\pmod 3$, therefore $f'(c)\not\equiv 0 \pmod p$.  
Hence there is some $d$ such that
$$
d^2+d+1\equiv 0 \pmod{p^2}
$$
$$\tag*{$\square$}$$

We now solve our main problem.  

Lemma. Let $p\equiv 1\pmod 3$ be a prime. Then there are integers $0<a,b<p$ such that
  $$
a^2+ab+b^2\equiv 0 \pmod{p^2}
$$

Proof. We have shown that there is some integer $d$ such that
$$
d^2+d+1 \equiv 0\pmod{p^2}
$$
Now by Thue's Lemma, since $\gcd(d,p^2)=1$, we can find $0<|a|,|b|<p$ such that
$$
db \equiv a \pmod{p^2}
$$
By rearranging the sign, we may assume that $b>0$ and $-p<a<p$. Now $a,b$ satisfies
$$
a^2+ab+b^2 \equiv d^2b^2+db^2+b^2 \equiv b^2(d^2+d+1)\equiv 0 \pmod{p^2}
$$
Therefore if $a> 0$ we have found our solution $0<a,b<p$. Otherwise, $a<0$ and we consider two cases ($-a\neq b$, otherwise $d\equiv -1\pmod{p^2}$ contradicts $d^2+d+1\equiv 0\pmod{p^2}$):
Case 1: $-a > b$
We use the equality
$$
a^2+ab+b^2 = (-a-b)^2 + (-a-b)b + b^2
$$
and obtain a solution $(a',b') = (-a-b,b)$. Since $0<-a<p$ and $0<b<p$, we have $-a-b< p$. By assumption $0 <-a-b$, therefore $0 < a',b' < p$ and we are done.  
Case 2: $-a < b$
We use another equality:
$$
a^2 + ab + b^2 = (-a)^2 + (-a)(a+b) + (a+b)^2
$$
and set the new solution as $(a',b') = (-a,a+b)$. Clearly $0 < a'=-a < p$. Since $-p < a < 0$ and $0 < b < p$, we get $a+b < p $. Finally, by assumption we have $0 < a+b$, therefore $0 < a+b=b' < p$. Hence we have a solution $0<a',b' < p$.  
$$\tag*{$\square$}$$
This completes the proof.  


Extra. Let $0<a,b<p$ and
  $$a^2+ab+b^2\equiv 0 \pmod{p^2}$$
  for some odd $p$. Then
  $$a^2+ab+b^2 = p^2$$ 

Proof. Since $0<a,b< p$, we have
$$
a^2 + ab + b^2 < 3p^2
$$
Morever, since $a^2+ab+b^2\equiv 0 \pmod{p^2}$, we must have $a^2+ab+b^2 = p^2$ or $2p^2$. If $a^2+ab+b^2=2p^2$ then
$$
a^2+ab+b^2 \equiv 0 \pmod 2
$$
There is no solution if $a$ or $b$ is odd, therefore both $a,b$ are even. But this means
$$
a^2+ab+b^2\equiv 0 \pmod 4
$$
Since $a^2+ab+b^2$ is also divisible by odd $p^2$, it is divisible by $4p^2$. This contradicts that $a^2+ab+b^2 < 3p^2$, therefore we must have
$$
a^2+ab+b^2 = p^2
$$
$$\tag*{$\square$}$$
