Alright /sci/ I'm doing some tutoring for some freshmen at my uni and I stumbled upon pic related.
How can we prove it by mathematical induction and just induction? I think it'll be a fair exercise.
>>8093523
a^2+b^2=c^2 is the best avenue of solving that equation.
>>8093523
[math]10\nmid(k^{2}+2k+2)(k+1)[/math] though, try n=3, you get 17*4 which is 68.
>>8093528
Incorrect.
k+1=60
60-1=k
K=59
>>8093530
what are you smoking? you can't prove [math](k^{2}+2k+2)(k+1)[/math] is a multiple of 10 because it's not always a multiple of 10.
>>8093530
in the case of k=3, [math](k^{2}+2k+2)(k+1)=17\times 2\times 2[/math]. You have your two twos, your 3 and 5 factors come from [math]k(k+1)(k+2)[/math] which equals [math]3 \times 4 \times 5[/math]
>>8093523
Either k and k+2 are even or (k+1)^2 is even so you have your factor of 4.
Assume k,k+1,k+2 are not divisible by 5. Then k=1 or 2 mod 5
k^2+2k+2=>
1^2+2+2=5=0 mod 5
2^2+4+2=10=0 mod 5
Thus you have your factor of 5 and so its 4*3*5*q for some q
>>8093535
Oh really buddy? I'll have you have you know that I'm a doctorate in astrophysics. This is how to solve the equation.
Cos=adj÷opp
Opp=cos×adj
59+1=60
60÷10=6
6^2=36
36=cos×adj
36=(k2+2k+2)(k+1)
Just use prime congruences and Hensel's lemma.
>>8093523
op just play with the factors and shit until you get
(k+1)^2 * ((k+1)^4 - 1) = k^2 * (k^4 - 1) * p
where p is some fucking other polynomial fuck
>>8093595
you do that and get a polynomial not entirely reducible, good luck proving it's a multiple of 60.
>>8093596
fuk u i have a triple minor in polynomials bitch
>>8093523
You need a lemma to prove this, and this lemma is another inductive proof.
The lemma you need is [6k^5 + 15k^4 + 20k^3 + 15k^2 + 4k] is divisible by 60. You can prove this by induction. Basis: 6+15+20+15+4=60*(1), Inductive step: 6(k+1)^5 + 15(k+1)^4 + 20(k+1)^3 + 15(k+1)^2 + 4(k+1) = [6k^5 + 15k^4 + 20k^3 + 15k^2 + 4k] + 30k^4 + 30k^2 + [120k^3 + 180k^2 + 180k + 60]. Clearly, the first term in brackets is divisible by 60 by the inductive hypothesis, and the second thing in brackets is obviously divisible by 60, but 30k^4 + 30k^2 requires more thought. 30k^4 + 30k^2 = 30(k^4 + k^2), and k^4 + k^2 is even for all k, so 30(k^4 + k^2) is divisible by 60, so [6k^5 + 15k^4 + 20k^3 + 15k^2 + 4k] is divisible by 60 for all natural k, and your lemma is proven.
>>8093820
and to apply the lemma, do this
Basis step: (1^2)*(1^4 - 1) = 0 = 60*0
Inductive step: [(k+1)^2]*[(k+1)^4 - 1] = (k^2)*(k^4 - 1) + (2k+1)*(k^4 - 1) + (k^2)*(4k^3 + 6k^2 + 4k +1) + (2k +1)*(4k^3 + 6k^2 + 4k + 1)
= (k^2)*(k^4 - 1) + 6k^5 + 15k^4 + 20k^3 + 15k^2 + 4k. By the inductive hypothesis, (k^2)*(k^4 - 1) is divisible by 60, and 6k^5 + 15k^4 + 20k^3 + 15k^2 + 4k is divisible by 60, for all natural k, by our lemma, so (k^2)*(k^4 - 1) is divisible by 60 for all natural k.
