I noticed that modulus of difference of any two odd numbers raised to even powers individually is always divisible by 4 and their sum is always congruent to 2(mod 4).
Example
|77^2 - 59^4| = 12111432 ≡ 0(mod 4)
53^2 + 15^6 = 11393434 ≡ 2(mod 4)
I haven't been able to prove that though. Can anyone provide a proof?
Has this been discussed before because I couldn't find any information about this anywhere on the internet of in MSE.
Thank you
$\text{To prove this, you have to know that the square of any odd number is always 1 modulo 4.}\\\text{The proof is shown below. }$ $$\text{Let 2n+1 be the odd number, }$$ $$(2n+1)^2$$ $$=4n^2+4n+1$$ $$=4(n^2+n)+1$$ $$\text{Therefore, }(2n+1)^2\equiv1(\operatorname{mod}4)$$ $\text{Because }(2n+1)^2\text{ is still a odd number,}\\\text{So }(2n+1)^n\equiv1(\operatorname{mod}4)\text{ when n is an even number. }\\\text{Now we can get into your finding.}\\\text{From above, we know that all odd numbers to even powers can be represented by }4n+1. $ $$\text{Let 4n+1 be the first number, 4k+1 be the second number, }$$ $$(4n+1)-(4k+1)=4(n-k)\equiv0(\operatorname{mod}4)$$ $$(4n+1)+(4k+1)=4(n+k)+2\equiv2(\operatorname{mod}4)$$ $\text{Now, these are proved. }$
