21st Mediterranean Mathematical Competition

Question

Prove that there exist infinitely many integers x, y, z for which the sum of the digits in the decimal representation of 4x^4 + y^4 - z^2 + 4xyz is at most 2.

Accepted Answer

This is an easy problem with many solutions. We rewrite

4 x^{4} + y^{4} - z^{2} + 4 x y z = (4 x^{4} + y^{4} + 4 x^{2} y^{2}) - (4 x^{2} y^{2} + z^{2} - 4 x y z) = (2 x^{2} + y^{2})^{2} - (2 x y - z)^{2} = (2 x^{2} + y^{2} - 2 x y + z) (2 x^{2} + y^{2} + 2 x y - z)

The two factors

A = 2 x^{2} + y^{2} - 2 x y + z

and

B = 2 x^{2} + y^{2} + 2 x y - z

add up to

A + B = 4 x^{2} + 2 y^{2}

. If we choose the values of

x

and

y

so that

A = 4 x^{2} = 4 \cdot 5^{2 n + 2}

and

B = 2 y^{2} = 2 \cdot 2^{2 n}

, then the product will become

A B = 2 \cdot 1 0^{2 n + 2}

and the sum of the digits will equal 2.

Summarizing, we pick an integer

n \geq 1

and set

x = 5^{n + 1}

and

y = 2^{n}

. The desired equation

A = 4 x^{2}

is equivalent to

2 x^{2} + y^{2} - 2 x y + z = 4 x^{2}

, and hence

z = 2 x^{2} + 2 x y - y^{2} = 2 \cdot 5^{2 n + 2} + 1 0^{n + 1} - 4^{n} .

Note that

z

indeed is a positive integer. As the described choice of

x, y, z

then yields

4 x^{4} + y^{4} - z^{2} + 4 x y z = 2 \cdot 1 0^{2 n + 2},

the proof is complete.