Six incredibly difficult maths puzzles that only the world’s elite can answer

Fermat oh Fermat

problem 2 - ibradiouf

so far we only know that ai are real numbers (not necessarilly integrers), also it's trivial to find positive real numbers ai so that their product equals 1 (if you take rationals for instance)

problem 2
we a2.a3.....an=1, we can rewrite this as an ! / a1! = an.an-1.an-2........a3.a2.a1 / a1, this is saying an factorial divide it by a1 factorial equal , as we know that a is a positive real integers thus 1 is the smallest result that a factorial could get, which make infer that a2.a3.....an=1 is only possible if and only if an=an-1=an-2=....=a2=1.
lets prove that (1+a2)^2(1+a3)^3....(1+an)^n > n^n , as we know that a2.a3.....an=1 then (1+a2)^2(1+a3)^3....(1+an)^n = (1+a)^n+n-1+....+2 = (1+a)^(((n^2+n)/2)-1).
so the equation becomes (1+a)^(((n^2+n)/2)-1) > (n^n), as a=1, we have
2^((n^2+n)/2)>2(n^n),by dividing both side by 1/n we have 2^((n+1)/2)>2n,hence for all n > 3 this equation is satisfied

@makingapunt
you said that 16a^4 – 8a^2 – 4abk + k + 1 = 0 and then simplify it to 4a(a^3 – 2a – bk) + k + 1 = 0, should not the simplification be 4a(4a^3 – 2a – bk) + k + 1=0.
and you said that A^4 – 2A^(2) + 1 / (A-1) = 4ab – 1 is just A^3 – A^2 – A + 1 = 4ab – 1, by multiplying the equation with a-1 you won't be back at the innate equation. otherwise i did enjoy your workout really ingenious.

to MakingAPunt there flawed points in your solution not sure they work
Q5:
Let's assume p prime number and it be p^h the highest power of p that divides 4ab−1 (a
n odd number). Hence, we infer that p^h is odd and it divides e (4a^2 − 1)^2 = (2a − 1)^2
(2a + 1)^2. Given that (2a + 1)-(2a − 1)=2 then their greatest common divisor is 2. Therefore, p^h can only divide either (2a − 1)^2 or (2a + 1)^2. It be 2k=h if h is even or 2k=h if h is odd then p^2k can only divide either (2a − 1)^2 or (2a + 1)^2 so p^k can only divide either (2a − 1) or (2a + 1). We have now that 2a ≡ ±1 mod p^k . If a ≡ 1 mod p^k in that case we have 0 ≡ 4ab − 1 =
2a · 2b − 1 ≡ 2b − 1 mod p^k from which we can infer that 2b ≡ 1 ≡ a mod p^k. If instead a ≡ −1 mod p^k in that case 0 ≡ 4ab − 1 = 2a · 2b − 1 ≡ −2b − 1 mod p^k from which we can infer that e 2b ≡ −1 ≡ a mod p^k. In any case we found that a ≡ b mod p^k.

QED