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Zorluk: OrtaOdd and Even Integers (Parity)

If xx, yy, and zz are integers such that x2+y2+z2x^2 + y^2 + z^2 is an odd integer, then the sum x+y+zx + y + z must be an odd integer.

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Cevap

The statement is True because the square of any integer preserves the parity of that integer, making the parity of x2+y2+z2x^2 + y^2 + z^2 identical to the parity of x+y+zx + y + z.
The statement is true because for all integers kk, the parity of k2k^2 is identical to the parity of kk. Consequently, the parity of the sum of squares x2+y2+z2x^2 + y^2 + z^2 is always identical to the parity of the sum x+y+zx + y + z. If the sum of squares is odd, the sum of the variables must also be odd.

Adım Adım Çözüm

1
Analyze the parity relationship between any integer kk and its square k2k^2.
If kk is even, k2k^2 is even; if kk is odd, k2k^2 is odd. (Note that zero is an even integer, and 02=00^2 = 0 is even).
Multiplying an even integer by itself yields an even number, and multiplying an odd integer by itself yields an odd number.
2
Relate the parity of x2+y2+z2x^2 + y^2 + z^2 to x+y+zx + y + z.
The expression x2+y2+z2x^2 + y^2 + z^2 has the exact same parity as x+y+zx + y + z.
Replacing each squared term with its base term does not change whether the sum is even or odd.
3
Apply the given condition to deduce the final parity.
Since x2+y2+z2x^2 + y^2 + z^2 is odd, x+y+zx + y + z must be odd.
Two expressions with identical parity must both be odd if one is given as odd.

Anahtar Kavram

Parity Invariance of Integer Powers
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