A Beautiful Number Theory Result | Iranian Mathematics Olympiad Second Round 1996

[@vinc17fr]

Much simpler: just factorize as follows, using ab = cd: a(a+b+c+d) = a²+cd+ac+ad = (a+c)(a+d). If a+b+c+d is a prime, then it divides a+c or a+d, which is impossible as being larger. (Note: the factorization can easily be found by first considering the case a = 1, then reintroducing a.)

[@WagesOfDestruction]

This is a much better proof.

[@dogandonmez5274]

Let us write a/c (which is equal to d/b) in lowest terms as x/y. Then a=mx, c=my, d=nx, b=ny for some (positive) natural numbers m,n. Then a+b+c+d=(m+n)(x+y), so it is not prime. (This is essentially same as the solutions given by Азирет Акматбеков and Я ЕРКАНАТ in simpler terms)

[@ravirajamadan]

This is such an elegant solution.

[@АзиретАкматбеков-й1м]

By using four number lemma, write a, b, c, d as: a=xy b=zw c=xz d=yw So, a+b+c+d equal to xy+zw+xz+yw x(y+z)+w(z+y) (z+y)(x+w)

[@Deathranger999]

If you happen to remember that lemma, yeah.

[@bait6652]

why would you have to remember it ? expand ab or cd to factors...then by equality, other pair has another combo. then factor as above.

[@Deathranger999]

@@bait6652 I mean yeah you don’t have to, but at that point it seems more complicated than either of the methods in the video.

[@jofx4051]

Oh well that's short and nice one

[@bait6652]

@@Deathranger999 mm i guess its not rigor but Exist k l m n : (a)(b)=(c)(d)=(kl)(mn) =(km)(lm)=(a)(b) N=sum=kl+mn+km+lm =Fac=(k+n)(m+l) both fac>= 2!=1 thus composite To make it rigor might require gcd or primefac...but seems unnecc

[@richardfredlund8846]

a*b = c* d ... let A*B:=a and P*Q:=b (where perhaps A and P =1) then if d=A*P and c=B*Q then a+b+c+d=A*B+P*Q+A*P+B*Q which factorizes as A*(B+P) + Q*(B+P) = (A+Q)*(B+P) where both factors are greater than 1.

[@ЯЕРКАНАТ]

If ab=cd then there exist four natural numbers that a=xy; b=zt; c=xz; d=yt a+b+c+d=xy+zt+xz+yt=(x+t)(y+z)

[@ЯЕРКАНАТ]

Sorry for my bad england

[@raphaelnej8387]

thanks for your good maths

[@luisisaurio]

This sounds like a really great moment to introduce a really cool technique that helps a lot in certain geo problems. If a/b=c/d then a/b=c/d=(a+c)/(b+d)=(a+b+c+d)/(b+d) that is if you have two fractions that are equivalent then it also is equivalent to the division of the sum of the numerators and the sum of the divisors. Proof Let a/b=c/d and x/y be the tiniest form of the fraction then a=x*m, b=y*m, c=x*n, d=y*n and (a+c)/(b+d)=x(m+n)/y(m+n)=x/y. and as I said it is really useful in geometry when you are working with ratios.

[@bookert2373]

1) multiply out both sides and use ab=cd to prove this equality: b(a+b+c+d) = (b+c)(b+d) 2) if none of a,b,c, or d =0, then each factor on right strictly exceeds b on left, so b(a+b+c+d), being divisible by, say b+d, must have some nontrivial sub factor in common with b+d, hence a+b+c+d isn’t prime.

[@leif1075]

What are you multiplying on both sides though??

[@bookert2373]

@@leif1075 just expand both sides; the terms bc, bd, and b^2 appear on both sides; what remains is the term ab in the left hand expansion and cd in the right, and those are given to be equal.

[@tmpqtyutmpqty4733]

That's a very nice proof.. we'll done

[@alinpopescu4147]

I didn't watch the video (yet) but I saw this problem before and here are 2 of my solutions: Solution 1: Suppose M=a+b+c+d is prime, then aM=a^2+ab+ac+ad=a^2+cd+ac+ad=(a+c)(a+d). Since M is prime and M|(a+c)(a+d) we get M|a+c or M|a+d but both are impossible because M=a+b+c+d>a+c and M=a+b+c+d>a+d; Thus we conclude that M is composite. Solution 2: ab=cd => a/d=c/b=u/v , where u/v is a irreducible fraction. Thus we could amplify the fraction u/v by m and n, to get a/d and c/d respectively. It follows that a=u*m; d=v*m; c=u*n; b=v*n. Therefore a+b+c+d= um+vn+un+vm=(u+v)(m+n) which is clearly composite. Challenge: Prove that if a number can be written as the sum of 2 squares in 2 different ways then this number is composite. i.e. Prove that for all positive integers a,b,c,d if (a,b)!=(c,d) and a^2+b^2=c^2+d^2=N then N is composite. Hint: Reduce the problem to a equation of the form AB=CD and make a substitution similar to the one in the second solution.

[@kubogi]

This is actually one of the problems in my high school entrance exam, and I solved it using polynomials (same idea but it sounds nicer)

[@stmmniko7836]

nice problem, keep us good work!

[@mrityunjaykumar4202]

if b either divides (d+b) or (c+b) then b is prime..if b is composite.. then there is a case that 'b' partially divides (d+b) and (c+b).. wbt?

[@toddtrimble2555]

A beautiful number theory result deserves a beautiful proof. Happily, such has been provided in comments.

[@mylokc4219]

so how do sum of 2 even number and 2 odd number is a prime ?

[@padraiggluck2980]

Toward the end, it seems to me that if p|(d+b)and q|(c+b) that we can only conclude that (d+b)/p and (c+b)/q are >=1. Maybe I’m misunderstanding something.

[@leodaric5447]

d+b > b > p and c+b > b > q so you know the inequality is strict

[@AndreyGoryainov-k7o]

2:37 (c+a) or (c+b) might be 0 => divide by p

[@Cookie-hq9kn]

Using basic parity: ab and cd are either even or odd If ab and cd are odd: a, b, c, and d are all odd because odd numbers multiplied are odd. However, odd + odd + odd + odd = even. (Short-handing for simplicity) If ab and cd are even: At least one factor must be even. The other factor can be either even or odd. However, even + odd + even + odd = even and even + even + even + even = even. Because a, b, c, and d are natural, their sum must be greater than or equal to 4 and there are no even primes greater than 4.

[@raphaelnej8387]

a = 3 b = 4 c = 2 d = 6 ab = bc a + b + c + d = 15 not even

[@dogandonmez5274]

Eve+even+even+odd is also possible for example: 4+1+2+2

[@floppitommi123]

Are a,b,c,d different numbers?

[@tmpqtyutmpqty4733]

So are we just going to ignore of how powerful this lemma seems?

[@morrispearl9981]

Would it make sense to make a parity argument. To sum to an odd number, the four numbers must have some even numbers and some odd numbers -- and there must be one odd number and three even number (the product equation could not work with three odds and one even). You can reduce the product equation mod 4, and show that there is no combination of three even numbers and one odd number that can make the equation true (mod 4).

[@lucaferrigno4767]

What about 2*2=4*1? Or 2*6=4*3?

[@petersievert6830]

not that easy after all, e.g. 2*4 = 8*1

[@UneFenetreSurLeMonde]

Because it's ≥ 2 then it's not a prime so 3 is > 2 so 3 is not prime ??

[@absolutezero9874]

Iranian Math Olympiad Second Round 1996: ab = cd Hence, a/c = d/b Let a/c = k, Hence, d/b = k Without loss of generality, Assume that a ≥ c and d ≥ b, k ≥ 1 Hence, a = kc --(1) d = kb --(2) Sub. (1) and (2) into the given expression, a + b + c + d = kc + b + c + kb = (b + c) + (kb + kc) = (b + c) + k(b + c) = (b + c)(1 + k) Since b, c ∈ ℕ, b ≥ 1, c ≥ 1 Hence, b + c ≥ 2 Since k ≥ 1, 1 + k ≥ 2 A prime number is divisible by 1 and itself only Since neither (b + c) nor (1 + k) can be equal to 1, (b + c)(1 + k) is never prime Hence, (a + b + c + d) is never prime

[@SuperYoonHo]

Thanks!

[@fdegreesx]

推，uesful and power method

[@parthibhayat]

Easiest number theory prob I have seen so far :)

[@hoangnguyennguyen6445]

the first is more lovely

[@ansumanc]

2:07 why can't p divide both c+a and c+b at the same time? It would still give 0 mod p

[@jakobr_]

The logical “or” includes the case of both statements being true. Either way, since p cannot divide c+a nor c+b, the case where p divides both also leads to a contradiction.

[@ansumanc]

@@jakobr_ ah that makes sense. Thanks

[@bait6652]

strange author check for 1*p factors rather then jumping straight to q*p.

[@jofx4051]

When u don't fluent in 100% math

[@JEREMYREOUVEN]

0 2 0 5 0x2 = 0x5 7 is prime

[@elliotachermann1487]

The numbers have to be in the natural set of numbers which means they have to be positive integers which means 0 is not allowed as it is not a positive integer

[@JEREMYREOUVEN]

@@elliotachermann1487 en.m.wikipedia.org/wiki/Natural_number 0 is a natural number.

[@JEREMYREOUVEN]

@@elliotachermann1487 and 0 is a positive integer. Just watch out before you argue something

[@peamutbubber]

Hey Google, is 0 a natural number? No 0 is not a natural number, natural numbers span integers 1 to infinity!

[@peamutbubber]

U r wrong, source: maths

[@ready1fire1aim1]

First ten numbers (0, 1, 2, 3,...9) First ten dimensions (0D, 1D, 2D, 3D...9D) Newton: "0 is contingent/not-necessary" 🚫 and "1-9 are necessary" 🚫 (this is the basis of Newton Calculus/Physics/Geometry/Logic). Leibniz: "0 is necessary" ✅ and "1-9 are contingent (on their predecessor)" ✅ (this is the basis of Leibniz Calculus/Physics/Geometry/Logic). [Info on Zero]: Is zero the most important number? Zero is the most important number in mathematics. Zero functions as a placeholder. Imagine a number, e.g., 5 and put as many zeroes behind it as you can think of. Zero drastically changes the value of the number from a mere 5 to 50, 500, 5000, 50000 and beyond. Which is the greatest whole number? There is no 'largest' whole number. Every whole number has an immediate predecessor, except 0. A decimal number or a fraction that falls between two whole numbers is not a whole number. Why is it impossible to divide by zero? The short answer is that 0 has no multiplicative inverse, and any attempt to define a real number as the multiplicative inverse of 0 would result in the contradiction 0 = 1. Is 0 a rational number? Yes, 0 is a rational number. Since we know, a rational number can be expressed as p/q, where p and q are integers and q is not equal to zero. Thus, we can express 0 as p/q, where p is equal to zero and q is an integer. Is 0 A whole number? The whole numbers are the numbers 0, 1, 2, 3, 4, and so on (the natural numbers and zero). Negative numbers are not considered "whole numbers." All natural numbers are whole numbers, but not all whole numbers are natural numbers since zero is a whole number but not a natural number. Why is 0 a good number? Zero helps us understand that we can use math to think about things that have no counterpart in a physical lived experience; imaginary numbers don't exist but are crucial to understanding electrical systems. Zero also helps us understand its antithesis, infinity, in all of its extreme weirdness. 🔘 ♾ ☯️ [Newton vs Leibniz]: Do you agree with Newton that "0 is contingent" and "1-9 are necessary"? Newton was a fraud and a moron. Clearly (shown) his logic is flawed at the fundamental level. (9 is contingent on 8, 8 is contingent on 7, and so on with the exception of 0; necessary) So why are we learning Newton's backwards Calculus/Physics/Geometry/Logic? [Newton vs Leibniz Calculus]: What is the difference between Newton and Leibniz calculus? Newton's calculus is about functions. Leibniz's calculus is about relations defined by constraints. In Newton's calculus, there is (what would now be called) a limit built into every operation. (1D-9D only; no correctly defined dimension above 3D) In Leibniz's calculus, the limit is a separate operation. (0D necessary and 1D-9D contingent)

[@pietergeerkens6324]

You are conflating the number 0 with the glyph "0".

[@ready1fire1aim1]

@@pietergeerkens6324 No. Just basic numbers and basic geometry.

[@ready1fire1aim1]

@@pietergeerkens6324 what glyph 0?

[@ready1fire1aim1]

@@pietergeerkens6324 0D is point 1D is line 2D is plane 3D is volume 4D is architecture 5D is design

[@alexandrezeddam7817]

What are you smoking dude? I really need to know