Linear combinations of squares

Clearly, $(a,b)=(0,0)$ is a solution. Assume $a,b\neq0$ . Let $a^2+6b^2=x^2 , b^2+6a^2=y^2$ . Adding them yields $7(a^2+b^2)=x^2+y^2$ . $7$ divides $x^2+y^2$ . The quadratic residue modulo 7 are 0,1,2,4. Therefore, both $x^2$ and $y^2$ are divisible by 7. Since 7 is a prime, $7|x$ and $7|y$ . Let $x=7x_1$ and $y=7y_1$ . $7(a^2+b^2)=(7x_1)^2+(7y_1)^2$ , $a^2+b^2=7(x_1^2+y_1^2)$ . This process will repeat indefinitely if $a$ and $b$ are not equal to 0 ( $a^2+b^2$ will approach infinity), a contradiction.

Let us assume that there exists a pair of integers (x,y) that satisfies these conditions and

$x^2 + 6y^2 = p^2$ ------(1)
$y^2 + 6x^2 = q^2$ _ ------(2)
where x, y, p, q $\in$ ${Z}$
(1)+(2) gives
_


$x^2 + y^2 + 6y^2 +6x^2 = p^2 + q^2$

If we add both expressions we have: 7(a^2+b^2)=c^2+d^2

Left side is a multiple of 7 and Right side must be a multiple of 7 for a solution to exist.

Both c and d can be expressed as 7k, 7k+1, 7k+2, 7k+3, 7k+4, 7k+5, 7k+6.

So, remainder when (c^2+d^2) is divided by 7 can be the same if one among the following is divided by 7:

0, (1+2^2), (1+3^2), (1+4^2), (1+5^2), (1+6^2)

(2^2+3^2), (2^2+4^2), (2^2+5^2), (2^2+6^2)

(3^2+4^2), (3^2+5^2), (3^2+6^2)

(4^2+5^2), (4^2+6^2)

(5^2+6^2)

None of the above except 0 is divisible by 7.

So, only one solution exist (a,b) = (0,0)

a^2 + 6b^2 = x^2 6a^2 + b^2 = y^2 a^2 + 6b^2 + 6a^2 + b^2 = 7a^2 + 7b^2 = x^2 + y^2 then mod 7 only one possible remainder is 0 so the solution is 1

Let

$a^2+6b^2=c^2\cdots (1)$ $b^2+6a^2=d^2\cdots(2)$

where $c$ and $d$ are integers.

If you add up $(1)$ and $(2)$ , you'll get:

$7(a^2+b^2)=c^2+d^2\cdots(3)$

So our question becomes, how many integer solutions are there for equation $(3)$ . That shouldn't be hard, because we have reduced the problem to another problem that we already solved :)

$(0,0,0,0)$ is the only integer solution.

If $a, b, c, d$ are all non-zero integers then $(3)$ has no solution.

Here's the proof.

Consider the solution $(a, b, c, d)$ such that $c^2+d^2$ is as small as it can be.

Notice that $7|(c^2+d^2)$ .

You can do a little case-work [consider the qudratc residues modulo $7$ ] and see that this means $7|c$ and $7|d$ .

Let $c=7k_1$ and $d=7k_2$ .

Then rewrite $(3)$ as

$7(a^2+b^2)=49({k_1}^2+{k_2}^2)$ $\Rightarrow 7({k_1}^2+{k_2}^2)=a^2+b^2$

Notice that this new equation is the same as $(3)$ . The only difference is that now $(a^2+b^2)$ acts as $(c^2+d^2)$ . It's not hard to see that $(a^2+b^2)<(c^2+d^2)$ . But that is a contradiction! Remember that we made $(c^2+d^2)$ as small as possible. But now we have a new $(c^2+d^2)$ that is less than that.

We arrived at this contradiction by assuming that the set of non-zero solutions to $(3)$ is non-empty. So that has to be false.

So, there are no more solutions. $(a, b)=(0,0)$ is the only solution.

And the answer is $\boxed{1}$ .

If $a^2 + 6b^2$ is a square then $b$ cannot be odd - if it was, $a^2 + 6b^2$ would be equivalent to 2 or 3 mod 4, which is impossible if the result is to be a square itself (i.e. 0 or 1 mod 4).

By the same argument if $6a^2 + b^2$ is a square then $a$ cannot be odd. Therefore we can assume that $a, b$ and hence the squared result, are all even.

Therefore the equations $a^2 + 6b^2 = x^2$ and $6a^2 + b^2 = y^2$ can both be divided through by 4 to create a two new expressions that have the same property, i.e. all three squares are even.

Repeated division by 4 must eventually make one or more of the squares odd, contradicting our assumption that the squares are all even, unless they are all zero.

Quite easily solved by infinite descent.

Denote the implicit squares in the problem as $a^2 + 6b^2 = n^2$ and $b^2 + 6a^2 = m^2.$ Adding the equations, we obtain $7 (a^2 + b^2) = n^2 + m^2.$ Now, consider $a+ib$ and $n+im$ as elements of Gaussian integers. Then the above equation tells us that the norms squared of these elements differ by factor of $7$ . Since unique factorization holds in Gaussian integers, this implies that $n+im$ is divisible by an element of norm $\sqrt{7}$ . But this is not possible since $7 = c^2 + d^2$ cannot hold. Both $c$ and $d$ would need to be smaller than $3$ and then the only possibilities are $0^2 + 1^2 = 1$ , $0^2 + 2^2 = 4$ and $1^2 + 2^2 = 5$ . So we have derived a contradiction whenever $a^2 + b^2$ is non-zero. There is however no problem with $a = b = 0$ , and so the answer is $\bf 1$ .

Let integer $k$ be a perfect square such that for any integer $n$ : $k = n^2$ .

Also, $a,b \in I$

From the problem,

$a^2 + 6b^2 = k_1$ ; $(1)$

$6a^2 + b^2 = k_2.$ ; $(2)$

We know that $k_i$ can only a perfect square integer as in:

$k_i$ $\in$ $[0, 1, 4, 9, 25, 36...]$

Since the problem do not specify that $k_1 \neq k_2,$ then we can assume that $k_1 = k_2$ , or $k_i=k_{i+1}$

If $k_1 = k_2,$

$a^2 + 6b^2 = 6a^2 + b^2$ ; $(3)$

then,

$5b^2 - 5a^2 = 0$ ; $(4)$

Essentially, this will yield a result:

$a=b$ , $(5)$

If we assume that a general perfect square for both eqn. 1 and eqn. 2 to be equal,

$k_i = k_{i+1}$ ; $(6)$

We can substitute eqn. 5 and 6 to eqn. 1, we get:

$a^2 + 6a^2 = k_i$ ; $(7)$

Applying the necessary algebraic operations,

$7a^2 = k_i$ ; $(8)$

Then,

$a^2 = \frac {k_i}{7}$ ; $(9)$

This yields to a general equation,

$a = b = \pm \sqrt{\frac {k_i}{7}}$ ; $(10)$

We know $k_i \in [0, 1, 4, 9, 25, 36...]$

By substituting values of $k_i$ in eqn. 10, we only get one integer ordered pair of $(a,b)$ , and that is when $k_i = 0$ . The rest are non-integer ordered pairs.

The only integer value of $(a,b)$ $\to$ $(0,0)$ .

So only $1$ ordered pair for can be a solution.

Let k,m be any 2 integers belonging to Z Let $a^2 + 6*b^2 = k^2$ ....(1) $\Rightarrow 6*a^2 + 36*b^2 = 6*k^2$ ......(2) Let $6*a^2 + b^2 = m^2$ ......(3) $\Rightarrow 36*a^2 + 6*b^2 = 6*m^2$ ....(4) Subtracting (3) from (2) and (1) from (4) we get $35*b^2=6*k^2-m^2$ $35*a^2=6*m^2-k^2$ Taking ratio we get $\Rightarrow (a^2/b^2) = (6*k^2-m^2 )/(6*m^2-k^2 )$ By Componendo-Dividendo $(a^2 + b^2)/(a^2 - b^2) = (5*m^2-5k^2)/(7*m^2-7k^2) \Rightarrow (a^2 + b^2)/(a^2 - b^2) = 5/7 \Rightarrow 7*a^2 + 7*b^2 = 5*a^2 - 5*b^2 \Rightarrow a^2 = -6*b^2$ No integer solution exists except for (0,0) So only 1 pair exists

Proof: From the problem we have that a^2+6b^2 and b^2+6a^2 equal some squares, lets name them c^2 and d^2 respectively. Thus we have a^2+6b^2=c^2 and b^2+6a^2=d^2.

We then add the two expressions, producing 7a^2+7b^2=c^2+d^2 =>7(a^2+b^2)=c^2+d^2 we now observe that c^2+d^2 must be divisible by 7.
c and d can be written as congruences mod 7. thus, 7n, 7n+1, 7n+2, 7n+3, 7n+4, 7n+5, 7n+6

since we can rewrite c and d in this way, the only thing required to check for divisibility by 7 is the remainder.

If you are wondering why we only need to check the remainders for divisibility by 7. look at this more general case.
take c=7n+x (0<=x<=6) c^2=(7n+x)^2= 49n^2+14nx+x^2 49n^2 is obiously divisible by 7, as well as 14nx. The only question arises from x^2, the remainder.

so we check these cases for divisibility. (0^2+0^2) (1^2+2^2), (1^2+3^2), (1^2+4^2), (1^2+5^2), (1^2+6^2) (2^2+3^2), (2^2+4^2), (2^2+5^2), (2^2+6^2) (3^2+4^2), (3^2+5^2), (3^2+6^2) (4^2+5^2), (4^2+6^2) (5^2+6^2) the only choice that works is (0^2+0^2) thus the only pair that satisfy this property is a,b=(0,0)