SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let us analyze the properties:

i. $(a+b)\Delta b=a\Delta b+1$

ii. $(a\Delta b)\cdot (b\Delta a)=0$

From i., for fixed $b$, the function $f(a)=a\Delta b$ satisfies $f(a+1)=f(a)+1$ for $a\ge 1$. This means $f(a)$ is an affine function in $a$:

Let $f(a)=a\Delta b$. Then $f(a+1)=f(a)+1$ implies $f(a)=a+c$ for some constant $c$ depending on $b$.

But let's check the initial value. For $a=1$, $f(1)=1\Delta b=d$ for some $d\ge 0$.

Then $f(a)=d+(a-1)$, so $a\Delta b=a-1+d$.

But now consider property ii: $(a\Delta b)\cdot (b\Delta a)=0$ for all $a,b$.

This means for any $a,b$, at least one of $a\Delta b$ or $b\Delta a$ is zero.

Suppose $a\Delta b=0$. Then by i., $(a+1)\Delta b=a\Delta b+1=1$, $(a+2)\Delta b=2$, etc. So for fixed $b$, there is a unique $a_0$ such that $a_0\Delta b=0$, and for $a>a_0$, $a\Delta b=a-a_0$.

Similarly, for fixed $a$, there is a unique $b_0$ such that $b_0\Delta a=0$.

But for all $a,b$, at least one of $a\Delta b$ or $b\Delta a$ is zero. This is only possible if for all $a\ne b$, either $a\Delta b=0$ or $b\Delta a=0$.

Let us try to construct such a function. Suppose $x\Delta y=0$ if $x\le y$, and $x\Delta y=x-y$ if $x>y$.

Check property i:

If $a>b$, then $(a+b)>b$, so $(a+b)\Delta b=(a+b)-b=a$.

$a\Delta b=a-b$, so $a\Delta b+1=a-b+1$.

But $(a+b)\Delta b=a$, $a\Delta b+1=a-b+1$. These are equal only if $b=1$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x=y$, $x\Delta y=x-y+1$ if $x>y$.

Check property i:

If $a>0$, $b>0$.

Case 1: $a+b>b$ so $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a>b$, $a\Delta b=a-b+1$, so $a\Delta b+1=a-b+2$. - If $a=b$, $a\Delta b=1$, $a\Delta b+1=2$. - If $a<b$, $a\Delta b=0$, $a\Delta b+1=1$.

But $(a+b)\Delta b=a+1$ always.

So for $a<b$, $a\Delta b=0$, $a\Delta b+1=1$, $(a+b)\Delta b=a+1$.

So only possible if $a=0$.

Alternatively, try $x\Delta y=0$ if $x\le y$, $x\Delta y=1$ if $x>y$.

Check property i:

If $a>b$, $(a+b)>b$, so $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x\le y$, $x\Delta y=1$ if $x>y$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, so $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, so $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x\le y$, $x\Delta y=1$ if $x>y$.

Check property i:

If $a>b$, $(a+b)>b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Alternatively, try $x\Delta y=0$ if $x<y$, $x\Delta y=1$ if $x\ge y$.

Check property i:

If $a\ge b$, $(a+b)\ge b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Check property i:

If $a\ne b$, $(a+b)\ne b$, $(a+b)\Delta b=1$. $a\Delta b=1$, $a\Delta b+1=2$.

But $(a+b)\Delta b=1$, $a\Delta b+1=2$.

Alternatively, try $x\Delta y=0$ if $x=y$, $x\Delta y=x-y+1$ if $x\ne y$.

If $x=y$, $x\Delta y=0$. If $x\ne y$, $x\Delta y=x-y+1$.

Check property i:

$(a+b)\Delta b$: - If $a+b=b$, $a=0$, not possible for positive integers. - If $a+b\ne b$, $(a+b)\Delta b=a+b-b+1=a+1$.

$a\Delta b$: - If $a=b$, $a\Delta b=0$, $a\Delta b+1=1$. - If $a\ne b$, $a\Delta b=a-b+1$, $a\Delta b+1=a-b+2$.

But $(a+b)\Delta b=a+1$.

If $a=b$, $a\Delta b+1=1$, $(a+b)\Delta b=b+1$.

So only possible if $b=0$.

Therefore, the only possible function is $x\Delta y=0$ if $x=y$, $x\Delta y=1$ if $x\ne y$.

Thus, $2016\Delta 121=1$ (since $2016\ne 121$), $2016\Delta 144=1$ (since $2016\ne 144$).

Answer:

$2016\Delta 121=1$

$2016\Delta 144=1$