Solution to Prove that if k and ℓ are posetive integers then k^2 − ℓ^2 can never … - Sikademy

Nov. 27, 2020

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Archangel Macsika

Prove that if k and ℓ are posetive integers then k^2 − ℓ^2 can never be equal to 2.

Solution

Proof

If k2 − ℓ2 has to be 2 then either both k and ℓ has to be even or both have to be odd.

If one is even and the other is odd then k2 − ℓ2 would be odd.

Now we solve case wise depending on whether both are odd or both are even.

Case 1: Both k and ℓ are even.

Then say k = 2m and ℓ = 2n when m and n are positive integers.

Then k2 − ℓ2 = (2m)2 − (2n)2

= 4m2 − 4n2

= 4(m2 − n2).

Now since (m2 − n2) is an integer and 4 times an integer cannot be 2 so k2 − ℓ2 cannot be 2 in this case.

Case 2: Both k and ℓ are odd.

Then say k = 2m + 1 and ℓ = 2n + 1 when m and n are positive integers.

Then k2 − ℓ2 = (2m + 1)−(2n + 1)2

= (4m2 + 4m + 1) − (4n2 + 4n + 1)

= 4(m2 + m − n2 - n).

Now since (m2 + m − n2 - n) is an integer and 4 times an integer cannot be 2 so k2 − ℓ2 cannot be 2 in this case.

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