Imperfection

Number Theory Level 4

Are there any three positive integers x x ; y y ; z z such that the following expression is a perfect square?

( x − 1 ) 2 + y 2 + ( z + 1 ) 2 + 2 ( y + 1 ) z + 2 ( z − 1 ) ( x + 1 ) + 2 x ( y − 1 ) \large \displaystyle (x - 1)^2 + y^2 + (z + 1)^2 + 2(y + 1)z + 2(z - 1)(x + 1) + 2x(y - 1)

Bonus

  • If there are solutions, finitely or infinitely, find all of them.

  • If there aren't any solutions, prove why.

No, there aren't any solutions. Yes, there are finite solutions. Yes, there are infinite solutions.

Thành Đạt Lê by Thành Đạt Lê , 17, Singapore

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2 solutions

If x = z x=z we get that the expression above is equal to { 2 z + y } 2 \left\{2z+y\right\}^2 so it is a perfect square for infinite cases .

5 Helpful 0 Interesting 1 Brilliant 0 Confused
Edwin Gray
Mar 8, 2019

Expanding the expression, we get x^2 + y^2 + z^2 + 2xy + 2xz + 2yz + 6z - 6x. = (x + y + z)^2 + 6(z - x). Let z = x, and A = 2x + y. The expression is A^2

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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