a delightfully symmetric number theory problem. 

Michael loves proofs by contradiction

We run into some problems. If 5y+1 is even, then 5y must be odd and therefore 2 cannot divide it, so 4 must divide 5y+1.

You are correct, 2 can divide y even though 4 doesn’t. Therefore you can’t assume that 2 doesn’t divide y, but that’s not what he did. He said that because 4 divides 5y(5y+1), and 4 has to divide at least one member of the product 5y(5y+1), and 4 doesn’t divide neither 5 nor y, 4 has to divide 5y+1. Hope I helped😊

@@omerhalaby2337 I think you're almost there, but the assertion "4 has to divide at least one member of the product 5y(5y+1)" requires the additional condition that gcd( 5, y, (5y+1) ) = 1. Fortunately, that's true.

we do always have that the gcd(a,x) =1 bx = -1 (mod a) -> bx +1 = k*a for some k in Z -> 1 = ka -bx that is Bezout's Identy therefore gcd(a,x) =1 In your counter example 21|(2*3)^2 + 2*3 but 21 ∤ (2*3) + 1 and that is required by construction

@@ruilopes6638 ya i specifically provided that example in which gcd(a,x)≠1 ,as micheal argument to do that was as x is in both term so . i provided counter to that argument that we cant do always. only if gcd is 1 then we can do . have a nyc day

no since 5y and 5y+1 are consecutive and cannot be both even.

@@AlcyonEldara Thanks, you are absolutely right.

The problem asks m and n satisfying the conditions for given a and b. Hence we can impose an extra condition as long as m and n indeed satisfying the conditions.

We're only asked to find some values for m and n that satisfy the three sets of conditions, and assuming x=y leads to set of values (namely those derived from x ≡ -1/b mod a and x ≡ -1/a mod b), so we have a solution. Or rather we have a set of solutions, since if x is a solution, then x+abk where k ∈ ℤ is also a solution, e.g. in the case of a=4, b=5, we find x=11 is a solution, but so is x=11+20=31. Check: m=124 and n=155 gives m^2+m = 15500 = 100n and n^2+n = 24180 = 195m, so they are good as well. In the case of a=4, b=5, we arrived at y ≡ 3 (mod 4) and x ≡ 1 (mod 5). So, we could try y=2x, but that would lead to 2x ≡ 3 (mod 4), not possible. No solutions there. Try x=2y, leading to y ≡ 3 (mod 4) and y ≡ 3 (mod 5), giving y ≡ 3 (mod 20), but taking m=12, n=15 fails to make m^2+m divisible by n. No solutions. Try y=3x, leading to x ≡ 1 (mod 4) and x ≡ 1 (mod 5), giving x ≡ 1 (mod 20), but that fails to make n^2+n divisible by m. No solutions. In fact, making y a multiple of x or vice-versa always results in numbers that are not "good". I wasn't able to prove a lack of solutions so far for arbitrary a, b such that gcd(a, b) = 1, and for relationships like jx = ky where j, k ∈ ℤ, but my conjecture is that no other solutions exist. Perhaps someone can provide the proof of that?

Standard notation for an inverse element b^-1 is a number, s.t. b^-1 • b = b • b^-1 = e (e - identity, for multiplication in modular arithmetic it is 1). So, for example: 4^-1 = 4 (mod 5) because 4 • 4 = 16 = 1 (mod 5) 2^-1 = 4 (mod 6) because 4 • 2 = 8 = 2 (mod 6) Btw, some elements may not have their inverses. There are no inverses for 2 (mod 4). Nevertheless b^-1 = 0 (mod a) sounds weird ‘cause b • 0 = 0 is true only for a = 0…

@@mrgold4678 I knew what means the inverse in the integers module n groups. A number not having inverse, as the zero, makes difficult to make deductions from, precisely, its inverse. The proof needs to stress that and a remake of that part. Better: a|bx -> bx=0(mod a), but from x=-b^(-1) (mod a) -> bx=-1 (mod a), showing a contradiction.

@@dodgsonlluis (Unless, of course, a = 1, but in this problem a must be strictly > 1.)

@@mrgold4678 You meant 2^-1 = 4 (mod 7) because 4 • 2 = 8 = 1 (mod 7)

b is invertible mod a iff gcd(a,b)=1

He sort of did. When he was looking for an x to satisfy both x ≡ 3 (mod 4) and x ≡ 1 (mod 5), he gave the answer x ≡ 11 (mod 20), and in general you would use CRT to find that.