Perfect Squares Pairs

It is clear that $a^2+3b > a^2$ and $b^2+3a > b^2$ . Since both of them are perfect squares, then there exist positive integers $k$ and $l$ for which $a^2+3b = (a+k)^2$ and $b^2+3a = (b+l)^2$ .

Case 1. $k \ge 2$ and $l \ge 2$ . We have $a^2+3b \ge a^2+4a+4$ and $b^2+3a \ge b^2+4b+4$ . Combining the two inequalities yields $0 \ge a+b+8$ , which is impossible. So, there is no solution.

Case 2. One of $k$ or $l$ equals 1, W.L.O.G. $k = 1$ . So, $a^2+3b = (a+k)^2 = a^2+2a+1 \Leftrightarrow 3b = 2a+1$ . Substituting this into the second equation, $b^2+3a = (b+l)^2 \Leftrightarrow 3\left(\frac{3b-1}{2}\right) = 2bl+l^2 \Leftrightarrow b(9-4l) = 3+2l^2$ . Since the right hand side of the last equality is always positive, then $9 > 4l$ which implies that $l = 1$ or $2$ . For the case $l = 1$ , we have $b = 1$ and $a = 1$ as well. For the case $l = 2$ , we have $b = 11$ and $a = 16$ . Hence, we have $3$ solutions: $(a,b) = (1,1), (11,16),(16,11)$ .

So, there are $3$ ordered pairs $(a,b)$ satisfying the problem.

Let $a^2+3b=(a+x)^2,b^2+3a=(b+y)^2[x,y \ge 1; x,y \in \mathbb{N}]$ . Adding the two equations we get $3a+3b=2ax+2by+x^2+y^2$ . If both $x,y \ge 2$ then $2ax+2by>3a+3b$ $\Rightarrow 2ax+2by+x^2+y^2>3a+3b$ which is a contradiction. So at least one of $x,y$ is $1$ . WLOG we assume $y=1$ .[Note: We are taking WLOG but we have to find ordered pair of solutions. So if $(p,q)$ is a possible pair where $p \neq q$ then $(q,p)$ will be a solution too.] Now $3a=2b+1 \Rightarrow b=\frac{3a-1}{2},3b=2ax+x^2$ . Plugging the first value in the second equation we get $9a-4ax=2x^2+3$ . If $x>2$ , then LHS becomes negative but RHS remains positive. So $x=1,2$ . Solving for $(x,y)=(1,1),(2,1)$ yields $(a,b)=(1,1),(11,16)$ . Since $11 \neq 16,(a,b)=(16,11)$ is another soluton. Hence we have three ordered pairs i.e. $(a,b)=(1,1),(11,16),(16,11)$ .

Without loss of generality, assume $0 < a \leq b$ . Hence, $b^2 < b^2+3a < b^2+4b+4=(b+2)^2$ so it must be that $b^2+3a=(b+1)^2$ if it is a perfect square. Therefore, $3a = 2b+1 \Rightarrow b = \frac{3a-1}{2} < 2a$ so $a^2 < a^2+3b < a^2 +6a < (a+3)^2 \Rightarrow a^2+3b = (a+1)^2$ or $(a+2)^2$ .

Case 1: $a^2+3b = (a+1)^2 = a^2+2a+1$ By simultaneously solving $3b=2a+1$ and $3a=2b+1$ from above, we have $a=b=1$ .

Case 2: $a^2+3b = (a+2)^2 = a^2+4a+4$ By simultaneously solving $3b=4a+4$ and $3a=2b+1$ from above, we have $a=11, b=16$ .

Therefore, there are a total of 3 ordered pairs, $(a,b) = (1,1),(11,16),(16,11)$

We can see that $x^2+3y < (x+2)^2$ or $y^2+3x < (y+2)$ because if both of inequalities are not true, we have $x^2+3y \ge (x+2)^2$ and $y^2+3x \ge (y+2)$ , then $x^2+y^2+3x+3y \ge x^2+y^2+4(x+y)+8$ leads to $0 \ge x+y+8$ contradiction with $x,y$ are positive integers.

Without loss of generality, we can assume that $x^2+3y < (x+2)^2$ , but it is easy to see $x^2 < x^2+3y$ , hence, $x^2 < x^2+3y < (x+2)^2$ and $x^2+3y$ is a perfect square. So $x^2+3y$ can only be $(x+1)^2$ or $3y=2x+1$ . Solve this Diophantine equation, we get $x=3k+1, y=2k+1$ with $k \in \mathbb{N}$ .

Continue, we have $y^2+3x = 4k^2+13k+4$ and with $k > 5$ , we have $(2k+3)^2 < 4k^2+13k+4 < (2k+4)^2$ , which implies that $y^2+3x$ can not be a perfect square. Therefore, $k \in \{0,1,2,3,4,5 \}$ .

Consider each of six cases, we obtain $k = 0$ and $k = 5$ : with $k = 0$ , we have $(x,y) = (1,1)$ and with $k = 5$ , we have $(x,y) = (11,16),(16,11)$ .

So in summary, there are 3 ordered pairs of positive integers satisfy all of the given conditions.

WLOG, let $a\le b$ (we will account for the ordered pairs after). Then note that $b^2<b^2+3a<b^2+4b+4=(b+2)^2$ , so for $b^2+3a$ to be a perfect square, $b^2+3a=(b+1)^2=b^2+2b+1\implies 3a=2b+1$ . Taking this equation modulo $3$ produces $b\equiv1\pmod{3}$ , so let $b=3c+1$ . Then $3a=6c+3\implies a=2c+1$ .

Therefore, $a^2+3b=(2c+1)^2+3(3c+1)=4c^2+13c+4$ . But note that $4c^2+13c+4<4c^2+16c+16=(2c+4)^2$ , and $(2c+3)^2=4c^2+12c+9<4c^2+13c+4$ if $c>5$ . Hence, for $c>5$ , $(2c+3)^2<a^2+3b<(2c+4)^2$ , so $a^2+3b$ can't be a perfect square.

If $c=5$ , $a=11,b=16$ , and $121+48=169$ and $256+33=289$ , so $a=11,b=16$ works.

If $c=4$ , $a=9,b=13$ , but $81+39=120$ , which isn't a perfect square.

If $c=3$ , $a=7,b=10$ , but $49+30=79$ , which isn't a perfect square.

If $c=2$ , $a=5,b=7$ , but $25+21=46$ , which isn't a perfect square.

If $c=1$ , $a=3,b=4$ , but $9+12=21$ , which isn't a perfect square.

If $c=0$ , $a=1,b=1$ , and $1+3=4$ , so $a=b=1$ works.

Consequently, the only ordered pairs are $(1,1),(11,16),(16,11)$ , for a total of $\boxed{3}$ ordered pairs.

since a^2 +3b is a square of some integer k and a,b are positive therefore a^2+3b=k^2(where k is an integer >a) therefore k=a+x (x is a positive integer). or, k^2=(a+x)^2 or, k^2=a^2+2ax+x^2 or, 2ax+x^2=k^2-a^2 or, 2ax+x^2=3b (since a^2+3b=k^2 therefore k^2-a^2=3b) or, 2ax-3b+x^2=0 ...............(1) again since b^2+3a is a square of some integer and a,b are positive therefore b^2+3a=m^2 (where m >b) therefore m=b+y(y is a positive integer). or, m^2=b^2+2by+y^2 or, 2by+y^2=m^2-b^2 or, 2by+y^2=3a(since b^2+3a=m^2 therefore 3a=m^2-b^2) or, -3a+2by+y^2=0 ..................(2) since a^2+3b and b^2+3a are both square number therefore equation (1)
and (2) holds. therefore, 2ax-3b+x^2=0 -3a+2by+y^2=0 this implies a/(-3y^2-2yx^2)=b/(-3x^2-2xy^2)=1/(4xy-9) i.e, a=(3y^2+2x^2y)/(9-4xy) and b=(2xy^2+3x^2)/(9-4xy) now a,b are positive integer and (3y^2+2x^2y),(3x^2+2xy^2) are positive because x and y are positive. therefore 9-4xy>0 therefore (x,y) can be (1,1) or (1,2), or (2,1) . when x=1,y=1 then a=1,b=1. when x=1,y=2 then a=11, b=16. when x=2,y=1 then a=16 and b=11. therefore (1,1);(11,16);(16,11) are the only three solutions of (a,b).

Set $a^2 + 3b = (a+c)^2$ and $b^2 + 3a = (b+d)^2$ , where $c$ and $d$ are positive integers. The equations are equivalent to $3b = 2ac + c^2$ and $3a = 2bd + d^2$ . Treating this as a system of linear equations in $a$ and $b$ , we get the solutions $a = \frac {2dc^2+3d^2} {9-4cd}$ and $b = \frac {2cd^2 + 3 c^2}{9-4cd}$ . Since $c$ and $d$ are positive, the numerators are positive. Thus, in order for $a$ and $b$ to be positive, we must have $9-4cd > 0$ . As such, $cd = 1$ or $cd = 2$ .

If $(c,d) = (1,1)$ , we get $(a, b) = (1,1)$ . If $(c,d)=(1,2)$ , we get $(a,b)=(16,11)$ . If $(c,d)=(2,1)$ , we get $(a,b) = (11,16)$ . Hence, there are 3 solutions in all.

let a^2+3b=n^2,and b^2+3a=z^2 add both eqn,(a^2)+(b^2)+3(a+b)=(n^2)+(z^2),as a,b&gt;=1 the minimum value of (n^2)+(z^2)=8. by am-gm inequality,(n^2+z^2)/2&gt;=sqrt(n^2 z^2) and equality holds good only when (n=z) or (n^2=z^2) thus maximum value of (n^2) (z^2)=16. thus from the above conditions we get (1,1) as the only solution.

Since,a^2+3 b is a perfect square then, a^2+3 b&gt;=0 a^2&gt;=-3b ........(1) and b^2+3 a&gt;=0 b^2&gt;=-3 a
a&gt;=b^2/-3 ..........(2)

Substituting (2) in (1), we get (-b^2/3)^2&gt;=-3*b b^3&gt;= -27 b&gt;= -3 .......,(3)

Substituting (3) in (2), we get a&gt;=-3

when putting b=3 in (1), we get a^2&gt;=-9, which is impossible

therefore, -3&lt;=b&lt;3 similarly, -3&lt;=a&lt;3

Hence, (1,1) is the only set which satisfy the condition that both a^2+3 b and b^2+3 a are perfect squares.

(1,1) is a solution which satisfies this equation. To satisfy the equation,both 'a' and 'b' should be odd numbers as difference between two square numbers which is a multiple of 5 should be odd. By using this fact a second pair can be obtained whose inverse is the third pair.

Let us call a=x and b=y

Since x and y are positive, we may write

x^2 + 3y = (x + a)^2, and y^2 + 3x = (y + b)^2

where a, b are positive integers.

Expanding, we find that the squared terms cancel, leaving the linear simultaneous equations (Simultaneous equations are a set of equations which have more than one value which has to be found).

3y = 2ax + a^2

3x = 2by + b^2

Solving, we obtain

x = (2a^2b + 3b^2)/(9 − 4ab)

y = (2b^2a + 3a^2)/(9 − 4ab)

Since a and b are positive, the numerator in each fraction will be positive. For the denominator to be positive, we must have ab = 1 or 2.

If (a,b) = (1,1), (1,2), (2,1), then, respectively, (x,y) = (1,1), (16,11), (11,16). Hence these are the 3 only solutions.

Now as a and b are positive integers, (a+1)^2 &gt; or = a^2+3b, which implies 2a+1 &gt; or = 3b. Also, (b+1)^2 &gt; or = b^2+3a will imply 2b+1 &gt; or = 3a. Combining these inequalities, a &gt; or = 1 and b &gt; or = 1. So the only possible positive integer solution is (1,1).

I have been able to show that the g.c.d of a and b is either 1 or 3.

Let $a^2 + 3b= x^2$ $b^2 + 3a= y^2$

Let (a, b)= d, so that a= ds and b= dr for some coprime integers and s.

So, $d^2*s^2 + 3dr= x^2$ Or, $d(ds + 3r)= x^2$

So, we can conclude that d divides x^2. However since x^2 is a perfect square, d must be present at least twice [or an even number of times] in the factorization of x^2.

But d is already present once in the factorization of x^2.

For d to be present twice in the factorization, d must divide ds + 3r

Or, d must divide 3r

Now, r= b/d.

So, d must divide 3b/d

Or, d^2 must divide 3b.

Now d^2 cannot divide b, the proof is given below.

On the contrary, assume that d^2 divides b, so that $b= pd^2$ for some integer p.

It has been shown that d divides x. So, x= dq, for some integer q.

So, $a= d^2 (q-3p)$

Therefore, d^2 divides a, and the g.c.d of a &amp; b is d^2, a contradiction [it can be possible only when d=1. Then if d not equal to 1, d=3, the proof of which is shown below].

Since $d^2$ cannot divide b, the gcd of d and b is d.

So, d must divide 3.

Thus d=1 or d=3.