Petar's equations

Subtracting the equations from each other, we have $xa = xb \to x(a-b) = 0 \to a=b$ or $x=0$ . This gives us three cases to consider.

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First, we consider $a = b \neq 0$ . This gives us $y \sin(\pi \sqrt{a^2 + b^2}) = y \sin(\pi a\sqrt{2})$ . Of course, this means that $xa = -y \sin(\pi a\sqrt{2}) \to x = -\frac{y \sin(\pi a\sqrt{2})}{a}$ , which means that we have infinitely many non-trivial solutions $(x,y)$ (setting $y \neq 0$ gives a non-zero real value for $x$ because no integer $a$ makes $\sin(\pi a\sqrt{2}) = 0$ ). This case gives us ten pairs for $(a,b)$ (namely, $(a,b) = (-5,-5),...,$ $(-1,-1),(1,1),...,(5,5)$ ) which satisfy the above conditions.

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Now we consider $a = b = 0$ . In this case, $\sin(\pi \sqrt{0}) = 0$ , and so we have $0x + 0y = 0$ , and so again we have an infinite number of non-trivial solutions. This gives us one pair for $(a,b)$ .

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Finally, if $x = 0, a \neq b$ , we have $0 + y \sin(\pi \sqrt{a^2 + b^2}) = 0$ and $y \neq 0$ (since $(x,y) = (0,0)$ is not a valid solution), which means that $\sin(\pi \sqrt{a^2 + b^2}) = 0 \to$ $\sqrt{a^2 + b^2} \in \mathbb{Z}$ . We need to find all the possible solutions to this with $\mid{a}\mid,\mid{b}\mid <6$ . If $a = 0$ , $b$ can be any integer except $0$ (since we assumed that $a \neq b$ ), which gives ten pairs $(a,b)$ . Similarly, if $b = 0$ , we get another ten pairs $(a,b)$ . Finally, we remind ourselves of the pythagorean triple $(3,4,5)$ , which tells us that $a = \pm 3, b = \pm 4$ (and vice versa) work, which gives us the final eight ordered pairs $(a,b)$ . Clearly, there are no other solutions on the given interval.

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Thus, we have a total of $10 + 1 + 10 + 10 + 8 = \fbox{39}$ ordered pairs $(a,b)$ satisfying the given conditions for which there exists a non-trivial solution $(x,y)$ to the given system of equations.

From the system,we get:

$xa = xb$ . So two cases follow:

Case 1: The solution $(x,y)$ is of form $(0,y) [y \neq 0]$ To avoid over counting we take $a$ & $b$ to be distinct.

So we arrive at: $y \sin (\pi \sqrt{a^2 + b^2}) = 0 \Rightarrow \sin (\pi \sqrt{a^2 + b^2}) = 0 \Rightarrow a^2 + b^2 = n^2$ for some $n$ .

$(a,b)$ can be $(0, \pm r)$ or $(\pm r , 0)$ for $r = 1,2,...,5$ & $(\pm 3,\pm 4)$ or $(\pm 4,\pm 3)$ , yielding a total of 28 solutions.

Case 2: if $a = b$ , then rearrangement gives: $x = - \frac{\sin (\pi \sqrt{2} a)}{a} y$ which has solution for $a = b = 0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5$ ,yielding total 11 solutions (0 is a solution obviously).

So there are total $39$ solutions.