# Kennard’s Ellipsoid Equation

Let us bring the search for math wizard to NCR. Another math quizzer named Kennard Ong Sychingping from De La Salle University-Manila wants to share a problem that can surely twist our brains. He is also in level 5 on his brilliant.org account in algebra.

I am a mathematics graduate, currently taking up Masters in the hopes of becoming a college or university professor one day. When I was in grade school, I was selected to join a mathematics training program inside our school. Ever since then, I have been passionate about mathematics. I consider to be lucky to have excelled in mathematics.” said Kennard.

Not to mention his math achievements in high school in Chiang Kai Shek College, he placed 2nd in the 19th Philippine Statistics Quiz Regional Finals (2010, NCR) and 1st Place in 2012 Sipnayan College Division.

Problem:

Find the number of integer solutions to the equation

$x^2+2y^2+98z^2=111\dots 111$(2014 times)

Solution:

Notice that $111\dots 111(2014 times)=11\dots 111 (2011 times)\cdot 1000+111$

Now we rewrite the equation as follows:

$x^2+2y^2+2z^2+96z^2=111\dots 111(2011 times)\cdot 1000+11$

Now we apply modulo 8 to both sides. This is the same as taking the remainder when dividing by 8.

Since both 96 and 1000 are divisible by 8, the resulting equation becomes:

$x^2+2y^2+2z^2\equiv 111\pmod{8}$

The remainder when 111 is divided by 8 is 7, so we can further reduce the equation to:

$x^2+2y^2+2z^2\equiv 7\pmod{8}$

Now the remainder when x is divided by 8 can either be 0, 1, 2, 3, 4, 5, 6, or 7. That is, x ≡ 0, 1, 2, 3, 4, 5, 6, or 7(mod 8).

If $x\equiv 0\pmod{8}$, then $x^2\equiv 0\pmod{8}$

If $x\equiv 1\pmod{8}$, then $x^2\equiv 1\pmod{8}$

If $x\equiv 2\pmod{8}$, then $x^2\equiv 4\pmod{8}$

If $x\equiv 3\pmod{8}$, then $x^2\equiv 9\pmod{8}$

Since 9 divided by 8 leaves a remainder of 1, then we have $x^2\equiv 1\pmod{8}$.

If $x\equiv 4\pmod{8}$, then $x^2\equiv 16\pmod{8}$. Since 16 is divisible by 8, then we have $x^2\equiv 0\pmod{8}$.

If $\equiv 5\pmod{8}$, then $x^2\equiv 25\pmod{8}$. Since 25 divided by 8 leaves a remainder of 1, then we have $x^2\equiv 1\pmod{8}$.

If $x\equiv 6\pmod{8}$, then $x^2\equiv 36\pmod{8}$.Since 36 divided by 8 leaves a remainder of 4, then we have $x^2\equiv 4\pmod{8}$.

If $x\equiv 7\pmod{8}$, then $x^2\equiv 49\pmod{8}$Since 49 divided by 8 leaves a remainder of 1, then we have $x^2\equiv 1\pmod{8}$.

Thus the only possible remainders for $x^2$ when divided by 8 are 0, 1, and 4.

That is, $x^2\equiv 0,1$,or $4\pmod{8}$. We can do the same to $y^2$ and $z^2$ to get $y^2\equiv 0, 1$, or $4\pmod{8}$ and $z^2\equiv 0, 1$, or $4\pmod{8}$

Now the equation that we have involves a $2y^2$ term and a $2z^2$ term. So for $y^2\equiv 0, 1$, or $4\pmod{8}$, we need to multiply both sides by 2.

If $y^2\equiv 0\pmod{8}$ then $2y^2\equiv 0\pmod{8}$

If $y^2\equiv 1\pmod{8}$ then $2y^2\equiv 2\pmod{8}$

If $y^2\equiv 4\pmod{8}$ then $2y^2\equiv 8\pmod{8}$

Since 8 is divisible by 8, this simplifies to $2y^2\equiv 0\pmod{8}$.

So $2y^2\equiv 0$ or 2(mod 8). We do the same thing to 4latex z^2\$ to get $2z^2\equiv 0$ or 2(mod 8).

We now have $x^2\equiv 0,1$, or 4(mod 8), $2y^2\equiv 0$ or 2(mod 8), and $2z^2\equiv 0$ or 2(mod 8). Adding the three equations, we check as to which combination will lead to the right-hand side of $x^2+2y^2+2z^2\equiv 7\pmod{8}$. We have:

0 + 0 + 0 ≡ 0(mod 8)

0 + 0 + 2 ≡ 2(mod 8)

0 + 2 + 0 ≡ 2(mod 8)

0 + 2 + 2 ≡ 4(mod 8)

1 + 0 + 0 ≡ 0(mod 8)

1 + 0 + 2 ≡ 3(mod 8)

1 + 2 + 0 ≡ 3(mod 8)

1 + 2 + 2 ≡ 5(mod 8)

4 + 0 + 0 ≡ 0(mod 8)

4 + 0 + 2 ≡ 6(mod 8)

4 + 2 + 0 ≡ 6(mod 8)

4 + 2 + 2 ≡ 8(mod 8), which can be reduced to 0(mod 8).

As none of the possible combinations leads to 7(mod 8), then no integer ordered triple (x, y, z) will satisfy the original equation. Thus the number of solutions is 0.

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

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