Two Solutions to International Mathematical Olympiad 1986 Problem 1

[@ManuelRuiz-xi7bt]

Let 13n - 1 = c² for some c in N. The only way c² = -1 (mod 13) is for c = +/-5 (mod 13). So let c = 13u +/- 5 for some u in N.
13 n = c² + 1
13 n = 13.13 u² +/- 13.10 +25 + 1
n = 13 u² +/- 10 u + 2
Discriminant = 100 - 104 < 0 -> no solutions for u, nor for n. Done.

[@moonlightcocktail]

Discriminant is 100 - 52(2 - n) for which solutions could exist. It is unreasonable to assume n = 0

[@ManuelRuiz-xi7bt]

@@moonlightcocktail You are right. I messed up.

[@mathslove89]

Thanks

[@TheColoursofMathematics]

Thanks a lot!

[@vaibhavcm7503]

Please explain the second proof again please.....

[@dominikwolski2274]

1. n must be odd, substitute n=2m+1
2. multiply all 3 numbers, you'll get polynomial in m, with constant term equal 38 and all other terms divisible by 4
3. 38 is 2 mod 4, all other terms will always be 0 mod 4, i.e. the product is always 2 mod 4
4. the product itself should be perfect square, so it must not be 2 mod 4
done

[@hydra147147]

Isn't the coefficient 48? After all, multiplying all numbers for n=1 gives 48 which is divisible by 4.

[@anjaneyasharma322]

Take n = 13 2 2 resptly

[@techysubham1939]

Crisp

[@gauravbharwan6377]

Bring some animation video

[@advaykumar9726]

First