MHB Let m and n be two integers. Prove that:

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The discussion focuses on proving that if the sum of the squares of two integers m and n is divisible by 4, then both integers must be even. A proof by contrapositive is suggested, assuming that m and n are odd, leading to the conclusion that m^2 + n^2 is not divisible by 4. The calculation shows that m^2 + n^2 results in a form that leaves a remainder of 2 when divided by 4. A participant points out that the contrapositive should consider the case where at least one of m or n is odd, suggesting a simplification of the proof. The conversation emphasizes the importance of correctly applying the contrapositive in mathematical proofs.
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Let m and n be two integers. Prove that if m2 + n2 is divisible by 4, then both m and n are even numbers

Hint: Prove contrapositiveAttempt:

Proof by Contrapositive. Assume m, n are odd numbers, showing that m^2 + n^2 is not divisible by 4.

let:
m= 2a + 1 (a,b are integers)
n=2b+1

m^2+n^2 = (2a+1)^2 + (2b+1)^2 = 4a^2 +4a + 1 + 4b^2 + 4b +1

let: 4(a^2 + a + b^2 + b) = 4q (q an integer)

m^2 + n^2 = 4q + 2

with 4q + 2 not divisible by 4 since 4 divides 4q + 2 with a remainder of 2.
==> if m and n are odd numbers, m^2 + n^2 is not divisible 4, which by contrapositive reasoning proves that if m^2 + n^2 is divisible by 4, then m and n are odd numbers.
 
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What is the negation of the statement

"They are both even" ?
 
KOO said:
Let m and n be two integers. Prove that if m2 + n2 is divisible by 4, then both m and n are even numbers

Hint: Prove contrapositiveAttempt:

Proof by Contrapositive. Assume m, n are odd numbers, showing that m^2 + n^2 is not divisible by 4.

let:
m= 2a + 1 (a,b are integers)
n=2b+1

m^2+n^2 = (2a+1)^2 + (2b+1)^2 = 4a^2 +4a + 1 + 4b^2 + 4b +1

let: 4(a^2 + a + b^2 + b) = 4q (q an integer)

m^2 + n^2 = 4q + 2

with 4q + 2 not divisible by 4 since 4 divides 4q + 2 with a remainder of 2.
==> if m and n are odd numbers, m^2 + n^2 is not divisible 4, which by contrapositive reasoning proves that if m^2 + n^2 is divisible by 4, then m and n are odd numbers.

Hi KOO! :)

Your proof looks fine...

But "contrapositive" in this case means that you start from NOT(m and n are even numbers).
That is the case if either m or n is odd, but not necessarily both.
Due to the interchangeability of m and n it will suffice if you pick m odd.

Perhaps you can redo your proof with just m odd and no information as yet about n?
 
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