Prove the following:

Question

𝐸𝑣𝑒𝑟𝑦 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑠𝑞𝑢𝑎𝑟𝑒 𝑖𝑠 𝑒𝑖𝑡ℎ𝑒𝑟 𝑎 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 4 𝑜𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 4𝑛 + 1 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑛.

(try proof by cases, since an integer is either even or odd, do one proof where 𝑛 is even, then do another one where 𝑛 is odd.)

Answer 1

well if N is even, then N = 2t for integer t

squaring: N^2 = 4t^2 which is a multiple of 4

if N is odd, then N= 2t+1 for integer t.

Squaring N^2 = 4t^2 + 4t + 1

= 4(t^2 + t) + 1

But t^2+t is also an integer by closure properties

of multiplication and addition over the integers

so in this case N^2 = 4K+1 for integer K=t^2+t

[end of proof]