Square your peas and queues

Assume, without loss of generality, that $p\geq q$ . Say $p^2 + pq + q^2 = a^2$ . Then $p^2 + 2pq + q^2 = (p+q)^2 = a^2 + pq$ . Using the difference of two squares factorization, $(p+q)^2-a^2=(p+q+a)(p+q-a) = pq$ . There are two cases.

Case 1 $\begin{cases} p+q+a=p\\ p+q-a=q \end{cases}$ This implies: $\begin{cases} q=-a\\ p=a \end{cases}$ So, $p=-q$ . But $p$ and $q$ are both positive, so this case yields no solutions.

Case 2 $\begin{cases} p+q+a=pq\\ p+q-a=1 \end{cases}$

Adding these two equations together, you get $\begin{aligned} 2p + 2q&=pq+1 \\ 2p -pq&=1-2q\\ p(2-q)&=1-2q \\ p&=\frac{1-2q}{2-q} = \frac{2q-1}{q-2} \end{aligned}$

Also, $2q-1 = 2(q-2)+3$ , so $q-2 \mid 2q-1 \implies q-2 \mid 3 \implies q-2 \in \{-3,-1,1,3\} \implies q \in \{1,3,5\}$ , as $q$ is positive. Three subcases:

• $q=1 \implies p = \frac{1}{-1} = -1$ , which is not positive.
• $q=3 \implies p = \frac{5}{1} = 5$ .
• $q=5 \implies p = \frac{9}{3} = 3$ , but $p\geq q$ , and $3\not\geq 5$ .

So, there is only one possibility, which is $p=5$ and $q=3$ . Then $a^2=p^2+pq+q^2=25+15+9=\boxed{49}$ .

So, write the equation into :

$p^2 + pq + q^2 = c^2$ , where $c$ is positive integer, then $(p+q)^2 - pq = c^2 \implies (p+q)^2 - c^2 = pq \implies (p+q+c)(p+q-c) = pq$ ,

And, we know that $RHS \ge 0 \implies LHS \ge 0 \implies (p+q-c) \ge 0$ . And, since $p+q + c \ge p + q$ . Since $(p+q-c) | pq$ , then

$p+q+c = pq \ldots \ldots (1)$

since $p+q+c \ge p$ and $q$ . So,

we also obtain $p + q - c = 1 \implies c = p + q - 1.\ldots \ldots (2)$

Then, from (1) and (2), we obtain : $p + q + (p + q - 1) = pq \implies (p - 2)(q - 2) = 3$

So, it's easy to see that $(p, q) = (3, 5)$ or $(5, 3)$

we obtain that $c^2 =p^2 + pq + q^2 = 3^2 + 3(5) + 5^2 = \boxed{49}$

When we see the $p^2+pq+q^2$ , we think of $(p+q)^2=p^2+2pq+q^2$ . Getting that into our equation, we get that $(p+q)^2-pq=k^2$ $(p+q)^2-k^2=pq$ $(p+q-k)(p+q+k)=pq$ Now, $pq$ has four factors. First, we try splitting $pq$ into $p$ and $q$ . This doesn't work, since it would mean that $k=\pm p=\mp q$ , but $p$ and $q$ are both positive. Therefore, $p+q-k=1$ $p+q+k=pq$ The sum of the equations gives $2p+2q=pq+1$ We can use SFFT (http://www.aops.com/Wiki/index.php/Simon's Favorite Factoring_Trick). To do so we move over all terms, and try to make a convenient factorization with a linear $p$ term and a linear $q$ term. $(p-2)(q-2)=3$ From this, $(p,q)$ is some permutation of $(3,5)$ , and so the answer is $3^2+3(5)+5^2=\boxed{49}$

Without loss of generality, let's assume m is positive, satisfy: p^2 + p.q +q^2 = m^2.

If we add p.q to both side we have, (p + q)^2 = m^2 + p.q. We can write this expression in a factorized form (p + q + m)(p + q - m) = p.q.

We consider the following cases: (a)(p + q + m) = p, and (p + q - m) = q.This is impossible

(b) (p + q + m) = q, and (p + q - m) = p.This is also impossible

(c) (p + q + m) = p.q, and (p + q - m) = 1. From (c) by addition we will have 2p + 2q = 1 + p.q, or 2p - 1 = q(p - 2), so that we get q = (2p -1)/(p - 2) or q = 2 + 3/(p - 2). The possible values of (p - 2) is 1, or 3. So the values of p are 3, or 5. And the values of q are 5, or 3 respectively. The value of m can be obtain by subtracting both equations. in (c), m = (p.q - 1)/2. So the value of m^2 is{ (p.q - 1)/2}^2 = 49

Let the number be $N = p^2 + p q + q^2 = (p + q)^2 - p q = (p + q - i)^2$ (for some integer $i < p,q$ )

So, $(p + q)^2 - p q = (p + q)^2 - 2 i (p+q) + i^2 \Rightarrow p q = i ( 2(p + q) - i)$

Here, $i$ divides RHS, so it must divide LHS i.e., $p$ or $q$ . But $i < p,q \Rightarrow i=1$

Put $i =1$ in the original relation to get: $p = \frac{2q - 1}{q - 2} = 2 + \frac{3}{q - 2}$ . Since $p$ is integral, $(q - 2)$ must be $\pm 1$ or $\pm 3$ . Only $q = 3 , 5$ give integral $p = 5 , 3$ respectively. For both, $(p + q - 1)^2 = 49$ .

If $p^2+pq+q^2=N^2$ , $pq=(p+q+N)(p+q-N)$ . We can assume without loss of generality that N is positive.

Since p and q are prime, either $p=p+q+N$ and $q=p+q-N$ or $pq=p+q+N$ and $1=p+q-N$ . If $p=p+q+N$ and $q=p+q-N$ , $0=q+N$ and $0=p-N$ , which cannot be true since $q>0$ . Thus, $pq=p+q+N$ and $1=p+q-N$ .

To simplify these equations, we cancel q and get $N=\frac{p^2-p+1}{p-2}=p+1+\frac{3}{p-2}$ . This also works for q: $N=p+1+\frac{3}{p-2}=q+1+\frac{3}{q-2}$ . Since N, p, and q are integers, $p-2,q-2$ must either be 1 or 3, meaning $p,q$ must either be 3 or 5. And if $p=q$ , $N^2=3p^2$ which is impossible. Thus, $(p,q)=(3,5)$ and $N=3^2+3(5)+5^2=49$ .

p^2 +pq +q^2=(p+q)^2-pq

Now because it has to be a perfect square let this be equal to x^2 for some natural number x

(p+q)^2-pq=x^2

(p+q)^2-x^2=pq

(p+q+x)(p+q-x)=pq

now because p and q are primes therefore

Case 1- (p+q+x)=p and (p+q-x)=q

this gives q=-x,p=x so we reject this case.

Case 2-(p+q+x)=q and (p+q-x)=q

this gives p=-x and q=x . so we reject this case.

Case 3- (p+q-x)=1 and (p+q+x)=pq

this case gives p+q=x+1, pq=2x+1

solving for p we get

p= x+1+√x^2-6x-3

because p is an integer x^2-6x-3 must be a perfect square

x^2-6x-3=(x-3)^2-12

It is easy to check that (x-3)^2 -12 is a perfect square for x=7

Now we just have to prove that this value of x is the greatest

For x>=7 (x-6)^2< x^2-6x-3<(x-3)^2

if x^2-6x-3=(x-5)^2 we get x=7

if x^2-6x-3=(x-4)^2 we get x=19/2 which is not a natural number

so only for x=7 x^2-6x-3 is a perfect square

also when x=7 p=5 and q=3 which are primes

So our answer is x^2=49 and the last three digits of 49 are 49

We factor over the Eisenstein integers. Let $\omega= \cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ . Then $p^2+pq+q^2 = (p-q\omega)(p-q\omega^2)$

We claim that the factors on the RHS are relatively prime. To see this, note that their greatest common divisor must divide their difference which is $q(\omega^2 - \omega)$ This is relatively prime to both factors however, so we must have $\gcd(p-q\omega, p-q\omega^2) = 1$

Since their product is a perfect square and they are coprime, there must exist positive integers $a$ and $b$ so that $p-q\omega = (a-b\omega)^2 = (a^2 - b^2) - (2ab+b^2)\omega$ where we have expanded and used the fact that $\omega^2 = -1 - \omega$ .

Now, this gives us $p =a^2-b^2=(a-b)(a+b)$ $q=2ab+b^2=b(2a+b)$

For $p$ to be a positive prime then, we require $a-b=1$ , so $a=b+1$ . Then $q=b(3b+1)$ implies $b=1$ (since $q$ is prime), so $a=2$ $b=1$

Then $p=3$ $q=5$ giving us $p^2+pq+q^2=49$

We re-write $p^2+pq+q^2$ as $(p+q)^2-pq$ . This must be the square of some integer $n$ which is clearly less than $p+q$ since $p,q>0$ . We express $n$ as $(p+q)-k$ , where $k$ is a positive integer.

Squaring $(p+q)-k$ gives us $(p+q)^2-2k(p+q)+k^2$ , which is what $(p+q)^2-pq$ is by definition. It follows that $pq = 2k(p+q) - k^2 = k(2p+2q-k)$ .

For the sake of contradiction, assume $k>1$ . Because $p,q$ are primes, either $p=k$ or $q=k$ and the other is $2p+2q-k$ . Without loss of generality, let $p=k$ . However, we now have $q = 2p + 2q - k = 2p + 2q - p =$ $p + 2q \to q = -p < 0$ , which is impossible because $p$ and $q$ are given to be positive.

Therefore, $k=1$ and $pq = 2p+2q-1$ . Rearranging, we have $pq-2p-2q=-1 \to (p-2)(q-2)-4=-1 \to$ $(p-2)(q-2)=3$ , which of course gives $p=3,q=5$ (or vice versa). Thus, our answer is $3^2+3\cdot5+5^2=\fbox{49}$ .

Here, p^2+pq+q^2=n^2. Then p^2+2pq+q^2=(p+q)^2=n^2+pq.

which gives (p+q)^2−n^2=(p+q+n)(p+q−n)=pq.

since p and q are primes ,

Two cases are possible.(taking p>q )

p+q+n=p and p+q−n=q which gives, q=-n (not possible) and p=n hence p+q+n=pq and p+q−n=1 Adding these two equations,we get

 2p+2q=pq-1
 pq + 2p + 2q + 4= 3
 (p-2)(q-2)=3

since q<p,
q - 2 = 1 and p - 2 = 3 is the only choice available. which gives q=3 and p=5 so p^2+pq+q^2=3^2+3(5)+5^2=49

Without loss of generality, let's assume m is positive, satisfy: p^2 + p.q +q^2 = m^2.

If we add p.q to both side we have, (p + q)^2 = m^2 + p.q. We can write this expression in a factorized form (p + q + m)(p + q - m) = p.q.

We consider the following cases: (a)(p + q + m) = p, and (p + q - m) = q.This is impossible

(b) (p + q + m) = q, and (p + q - m) = p.This is also impossible

(c) (p + q + m) = p.q, and (p + q - m) = 1. From (c) by addition we will have 2p + 2q = 1 + p.q, or 2p - 1 = q(p - 2), so that we get q = (2p -1)/(p - 2) or q = 2 + 3/(p - 2). The possible values of (p - 2) is 1, or 3. So the values of p are 3, or 5. And the values of q are 5, or 3 respectively. The value of m can be obtain by subtracting both equations. in (c), m = (p.q - 1)/2. So the value of m^2 is{ (p.q - 1)/2}^2 = 49

let, p^2+pq+q^2=m^2 then, (p+q)^2−pq=m^2 \Rightarrow (p+q)^2−m^2=pq \Rightarrow (p+q+m)(p+q−m)=pq we know, that p,q,m \geq 0, so (p+q+m)>p [(p+q+m) \neq p], (p+q+m)>q[(p+q+m) \neq q] and (p+q+m)>(p+q-m)[p,q>0] and these implies (p+q+m)=pq[we know that p,q are primes] and (p+q-m)=1 So, 2(p+q)=pq+1 \Rightarrow pq-2p-2q+1=0 \Rightarrow (p-2)(q-2)-3=0 \Rightarrow (p-2)(q-2)=3. this is symmetric on p and q, so either (p-2)=3 \Rightarrow p=5 and (q-2)=1 \Rightarrow q=3 Or, (p-2)=1 \Rightarrow p=3 and (q-2)=3 \Rightarrow q=5 So, (p,q)=(3,5) and (5,3) the given equation is symmetric on p and q, so, the value of the equation is (3^2 + 3 \times 5 + 5^2)=49[answer]

Without loss of generality, let $p \ge q$

We let $p^2+pq+q^2=r^2$ . Then $p^2+2pq+q^2=r^2+pq$ , so $(p+q)^2=r^2+pq$ , and $(p+q)^2-r^2=pq$ . Hence $(p+q+r)(p+q-r)=pq$ . Since $p$ and $q$ are primes, we have 2 cases: $p+q+r=pq$ and $p+q-r=1$ (this is because $p+q+r>p+q-r$ , or $p+q+r=p$ and $p+q-r=q$ (this is because $p+q+r>p+q-r$ and $p>q$ .

Case 1: $p+q+r=pq$ and $p+q-r=1$ Then $r=p+q-1$ , so $p+q+p+q-1=pq$ . Thus $2p+2q-1=pq$ . Hence $pq-2p-2q+1=0$ , and $(p-2)(q-2)-3=0$ . Thus $(p-2)(q-2)=3$ . Since $p \ge 2$ and $q \ge 2$ , both $p-2$ and $q-2$ are positive. Hence $p-2=3$ and $q-2=1$ , meaning that $p=5$ and $q=3$ . Hence, $p^2+pq+q^2=(5)^2+(5)(3)+(3)^2=25+15+9=49$ .

Case 2: $p+q+r=p$ and $p+q-r=q$ From the first equation we get $q+r=0$ . Thus $q=-r$ . From the 2nd equation we get $p-r=0$ , so $p=r$ . Hence $-p=-r$ , so $-p=q$ . However, $p$ and $q$ are positive so this is not possible.

Thus, the largest perfect square that can be expressed in the form $p^2+pq+q^2$ , where p and q are (positive) prime numbers is $49$ , so it's last 3 digits is $\boxed{049}$

Let p^2+pq+q^2=a^2. Then p^2+2pq+q^2=(p+q)^2=a^2+pq.

which gives (p+q)^2−a^2=(p+q+a)(p+q−a)=pq.

since p and q are primes , L.H.S has exactly two prime factors.

Two cases are possible.(taking p>q )

      p+q+a=p and  p+q−a=q
      which gives, q=-a (not possible) and  p=a

hence p+q+a=pq and p+q−a=1 Adding these two equations,we get

      2p+2q=pq-1
      pq + 2p + 2q + 4= 3
      (p-2)(q-2)=3 (a priime number)
      since q<p,      q - 2 = 1 and p - 2 = 3 is the only choice available
      which gives q=3 and p=5

*So,there is only one possible such number=3^2 + 3 5 +5^2=49. **

In the problem it's given that: p2+pq+q2, now completing the square (p+q)2=p2+2pq+q2. and getting that into the problem, the following is obtained:

(p+q)2−pq=a2

(p+q)2−a2=pq Factoring it as the difference of 2 squares......... (p+q−a)(p+q+a)=pq Since p,q are primes , pq has four factors. First, pq is split into p and q. It would mean that 2(p+q)=p+q , p+q=0 but p and q are both positive. So this case can't be taken into account. So, p+q−a=1 p+q+a=pq (Sine p, q, a are all positive their sum is always > 1, and p,q are themselves > 1 ;so p+q+a is taken to be = pq). Adding the equations............ 2p+2q=pq+1

Doing a little algebra ( this is often a standard technique which results from an effective factorization by taking pq to L.H.S. and taking p common then adjusting the remaining )

(p−2)(q−2)=3

From here possible values are: p-2=1 and q-2=3 or the reverse that is p-2=3 and q-2=1

(note that taking other values viz p-2=-1 and q-2=-3 would yield negative solutions so this case is avoided)

the equations give p=5,q=3 or p=3,q=5......putting this values in : a= pq-p-q gives a=7 and squarring it gives a as 49. So the answer is 49.

$p^2+pq+q^2=(p+q)^2-pq=k^2$ where $k^2$ is the perfect square. Rearranging, we get: $(p+q)^2-k^2=pq \Longrightarrow (p+q+k)(p+q-k)=pq$ We have case 1: p=q. Checking, we can clearly see that this is impossible case 2: $p+q+k=pq$ and $p+q-k=1$ (It can't be the other way around) By elimination, we get that $p+q+(p+q-1)=pq \Longrightarrow (p-2)(p-2)=3$ We see that $(3,5)$ is a valid solution, so we plug that in to get $\boxed{49}$

p^2 + pq +q^2 = k^2 ,

or, (p+q)^2 - pq =k^2

or, (p+q-k)(p+q+k) = pq

which gives, either

(i) p+q-k = p and p+q-k =q

(ii) p+q-k=q and p+q+k=p

(iii) p+q-k=pq and p+q+k=1

(iv) p+q-k =1 and p+q+k =pq

from (i) we get p = -q not possible as p and q are +ve primes.

from (ii) we get p = -q not possible as p and q are +ve primes.

eliminating k from (iii) and (iv) we get 2(p+q) = pq+1

or (p-2)(q-2)=3.

solving we get either p=3 and q=5, or p=5 and q=3.

substituting in p^2 +pq +q^2 = (3)^2 + (3)(5) + (5)^2 = 49 = (7)^2.

threrefore, last three digits are 049 or 49.

As far as I can tell, the only value possible is 49 (so 049). Call this perfect square a^2 with a>0 Comparing to the expansion of (p+q)^2, pq = (p+q)^2-a^2= (p+q+a)(p+q-a) Since both terms on the Rhs are integers, we can consider the factors of pq which, since p and q are prime are 1,p,q,pq and p+q+a and p+q-a must then fit into this set. Since p,q,a are >0, p+q+a cannot be equal to p or q so these factors must be pq and 1 respectively. The lesser is p+q-a=1 => a=p+q-1 The greater is pq= p+q +a = 2p+2q-1 So q(p-2)=2p-1, q= (2p-1)/(p-2) Since q is an integer, q-2 is also an integer which is (2p-1-2p--4)/(p-2)= 3/(p-2) So p-2 is a factor of 3, which is either 1 or 3. P is either 3 or 5, q is either 5 or 3 (in pairs of 3 and 5). This gives the expression given as 25+15+9=49 and a=7 Let me know if I've missed anything- but unless there's a mistake this is the only such square.

Call this perfect square a^2 with a>0

Comparing to the expansion of (p+q)^2, pq = (p+q)^2-a^2= (p+q+a)(p+q-a)

Since both terms on the Rhs are integers, we can consider the factors of pq which, since p and q are prime are 1,p,q,pq and p+q+a and p+q-a must then fit into this set.

Since p,q,a are >0, p+q+a cannot be equal to p or q so these factors must be pq and 1 respectively.

The lesser is p+q-a=1 => a=p+q-1

The greater is pq= p+q +a = 2p+2q-1

So q(p-2)=2p-1, q= (2p-1)/(p-2)

Since q is an integer, q-2 is also an integer which is (2p-1-2p--4)/(p-2)= 3/(p-2)

So p-2 is a factor of 3, which is either 1 or 3. P is either 3 or 5, q is either 5 or 3 (in pairs of 3 and 5).

This gives the expression given as 25+15+9=49 and a=7

It's possible when $(p+q)^{2}-(p+q-k)^{2} = pq$

=> $2kp+2kq-k^{2}$ = pq

=> p(q-2k) = $2kq-k^{2}$

=> p(q-2k) = $2k(q-2k)+3k^{2}$

=> p = 2k+ $\frac{3k^{2}}{q-2k}$

so q-2k = 3 or q-2k = 1, in both case (p,q,k) = (5,3,1)

So answer = $3^{2}+\times{3}{5}+5^{2}$ = 49

If p^2+pq+q^2=n^2 then pq=(p+q)^2-n^2=(p+q+n)(p+q-n) Since p, q are positive primes, we must have p+q+n=pq, p+q-n=1, so 2 (p+q)=pq+1, so (p-2)(q-2)=3. Thus {p, q} = {5, 3}, so n=pq-p-q=7, so n^2=49

$p^2+pq+q^2=k^2 \rightarrow (p+q-k)(p+q+k)=p$ . Assume WLOG $p \geq q$ . Since $p$ & $q$ are both primes and $p+q+k \geq p+q-k$ we have that either $p+q-k=p$ & $p+q+k=q$ or $p+q-k=1$ & $p+q+k=pq$ . The first case gives us that $p=-k$ which is a contradiction as both $p$ and $k$ are positive integers. The second case gives us that $p+q+k=pq \rightarrow (p-1)(q-1)=k+1$ and $p+q-k=1 \rightarrow p+q=k+1$ . Solving for $q-1$ , we get that $(p-1)(q-1)=p+q \rightarrow q-1=\frac{p+1}{p-2}$ . Thus $(p-1)(q-1)=k+1 \rightarrow (p-1)(\frac{p+1}{p-2})=k+1$ . Since $p-1$ and $p-2$ differ by one, they are relatively prime. Thus $p-2|p+1$ . Since $p-2$ must divide $p+1$ and they differ by $3$ and $p-2\ge 0$ , $p-2$ must equal $3$ . This means that $p=5$ and $k^2=49$ .

Let $p^{2} + q^{2} + pq$ = $n^{2}$ -------( $\ast$ ) for some positive integer, $n$ .

From ---( $\ast$ ) we manipulate as ${(p + q)}^{2} - pq$ = $n^{2}$

or ${(p + q)}^{2} - n^{2}$ = $pq$

or $(p + q + n)$ $(p + q - n)$ = $pq$

Now, since $p$ and $q$ are primes cases are:

(i) $(p + q + n)$ = $p$ and $(p + q - n)$ = $q$ , which implies $p + q$ = $0$ which is impossible,

(ii) $(p + q + n)$ = $q$ and $(p + q - n)$ = $p$ , which is similar to (i), so ruled out,

(iii) $(p + q + n)$ = $pq$ and $(p + q - n)$ = $1$ $\big($ this is equivalent to saying that $(p + q + n)$ = $1$ and $(p + q - n)$ = $pq$ $\big)$ ,adding the two we get

$2p + 2q$ = $1 + pq$

or $(p - 2)$ $(q - 2)$ = $3$

or $(p - 2)$ = $3$ and $(q - 2)$ = $1$

or $(p,q)$ = $(5,3)$ and vice-versa

Plugging in $(p,q)$ = $(5,3)$ in eqn-( $\ast$ ), we find $n^{2}$ = $5^{2} + 5$ $\times$ $3 + 3^{2}$ = $5$ $\times$ $(5 + 3)$ + $3^{2}$ = $40$ + $9$ = $\boxed{49}$ which is our answer.

Suppose $p^{2}+pq+q^{2}=k^{2}$ , wheere $k$ is positive integer,then $(p+q)^{2} -pq =k^{2} \\$ $(p+q)^{2}-k^{2} =pq \rightarrow (p+q+k)(p+q-k)=pq \\$ , $\\$ we know that $RHS\geq0$ and $LHS\geq0 \\$ $(p+q-k)\geq0$ And,since $p+q+k \geq p+q\mid pq$ ,then $p+q+k=pq$ since $p+q+k\geq p$ and $q$ .So, we can obtain $p+q-k=1 \rightarrow k=p+q-1\\$ from that we can get $p+q+(p+q-1)=pq \\$ $(p-2)(q-2)=3\\$ $(p.q)=(5,3) or (3,5)\\$ we can obtain that $p^{2}+q^{2}+pq=3^{2}+5^{2}+3.5=49$

Let said perfect square be represented by $a^2$ , where $a$ is a positive integer. Then we have that $p^2+pq+q^2 = (p+q)^2-pq = a^2 \implies (p+q-a)^2 = pq$ Since $pq$ can only be factored in two ways ( $p, q$ or $1, pq$ ) due to $pq$ being a product of only two primes, we have two possible cases:

Case 1: $p+q-a = p, p+q+a=q$ (assuming wlog that $p<q$ ) Simplifying leads to the reduced equations $q=a$ and $p=-a$ , a contradiction, since $p$ must be positive. So this case cannot happen.

Case 2: $p+q-a = 1, p+q+a=pq$ From the first equation we have $p+q=a+1$ . Working with the second equation, we get that $pq-p-q=a \implies (p-1)(q-1) = pq-p-q+1 = a+1$ However, since $p+q=a+1$ , we have $p+q = (p-1)(q-1) \implies p+q = pq-p-q+1 \implies pq-2p-2q+4=3 \implies (p-2)(q-2) = 3$ Since the only way to factor 3 is into $1, 3$ , we must have $p-2=3, q-2=1$ (or the other way around), which results in $p=5, q=3$ . Then $a^2=p^2+pq+q^2 = 25+15+9 = \boxed{049}$ .

p^2+q^2+pq = (m^2+k^2+mk)2 = (m^2-k^2)2+(k^2+2mk)2+(k^2+2mk)(m^2-k^2) p = m^2 - k^2 and q = k^2 + 2mk since q is a prime ,k = 1 and hence q = 2m+1 and p = m2-1 since p is also prime p = (m-1)(m+1) => m-1=1 i.e m=2 hence we got p=3 and q=5 as only solution thus p^2+q^2+pq = 49 hence last three digits of perfect square is 049

Without loss of generality assume that $p\leq q$ .

Let $p^2+pq+q^2=(p+k)^2$ , where $k$ is a positive integer. Now if $k\geq q$ then $p^2+pq+q^2=(p+k)^2\geq(p+q)^2=p^2+2pq+q^2$ , or $pq\leq0$ , contradiction. Hence $0<k<q$ , so $q\nmid k$ .

From $p^2+pq+q^2=(p+k)^2$ we get $(p+q)q=(2p+k)k$ . Since $q\nmid k$ , we have $q|2p+k$ . But $0<2p+k<2q+q=3q$ , so either $2p+k=q$ or $2p+k=2q$ .

If $2p+k=q$ then $p^2+pq+q^2=(p+k)^2=(q-p)^2=p^2-2pq+q^2,$ so $pq=0$ , contradiction.

If $2p+k=2q$ then $p^2+pq+q^2=(p+k)^2=(2q-p)^2=p^2-4pq+4q^2,$ so $4q^2-4pq=pq+q^2$ . Factoring out $q$ and rearranging terms, we get $3q=5p$ . Thus $3|p$ and $5|q$ , so $(p,q)=(3,5)$ , and thus $p^2+pq+q^2=9+15+25=49$ .

Since 49 is the only square of the form $p^2+pq+q^2$ with $p,q$ prime, the answer is $\boxed{49}$ .

Let $x$ be a positive integer. Then we have the equation $p^2+pq+q^2 = x^2,$ or $(p+q)^2-pq = x^2.$ Rearranging this, we get $(p+q)^2-x^2 = pq,$ and by difference of squares, $(p+q+x)(p+q-x) = pq.$ Without loss of generality, let $p \ge q.$ Also notice that $p+q+x > p+q-x$ because $x$ is positive. Because $p$ and $q$ are primes, the only possible factor pairs are $p, q$ and $1, pq,$ so we have the cases $p+q+x = p,$ $p+q-x = q,$ and $p+q+x = pq,$ $p+q-x=1.$ Taking the first equation of the first case, we see that $q+x = 0,$ which is impossible because both $q$ and $x$ are positive.

Taking the second case, we add the two given equations to get $2p+2q = pq+1$ $pq - 2p - 2q +1=0$ $(p-2)(q-2) = 3$ So, $p-2=3$ and $q-2=1,$ hence $p = 5, q = 3.$ Substituting into the equation $p^2+pq+q^2=x^2,$ we get $x^2 = 5^2 + 5(3) + 3^2 = 49.$ Thus, the answer is $\boxed{49}.$

p^2+pq+q^2 is a perfect square the it must be equal to some positive integer k^2 then p^2+pq+q^2=k^2 or (p+q)^2-pq=k^2 or (p+q)^2-k^2=pq (p+q+k)(p+q-k)=pq now two cases arise case-1 p+q+k=p p+q-k=q on solving we get p=-q which is not possible as p and q are +ve case-2 p+q+k=pq p+q-k=1 now we eliminate k from equations we get p+q+(p+q-1)=pq =>(p−2)(q−2)=3 (p,q)=(3,5) or (5,3) so p^2+pq+q^2=32+3(5)+52=49

p^2+pq+q^2 is a perfect square the it must be equal to some positive integer k^2 then p^2+pq+q^2=k^2 or (p+q)^2-pq=k^2 or (p+q)^2-k^2=pq (p+q+k)(p+q-k)=pq now two cases arise case-1 p+q+k=p p+q-k=q on solving we get p=-q which is not possible as p and q are +ve case-2 p+q+k=pq p+q-k=1 now we eliminate k from equations we get p+q+(p+q-1)=pq =>(p−2)(q−2)=3 (p,q)=(3,5) or (5,3) so p^2+pq+q^2=32+3(5)+52=49 key technique:-simple algebra manipulation and lil bit knowledge about primes and factorization