A cool number theory problem!

[@Bodyknock]

2:32 Doing these square tables simplifies if you instead look at x in -3, -2, -1, 0, 1, 2 or 3. (I.e. 4=-3 mod 7, 5=-2 mod 7, etc). When you write it out this way it becomes obvious that squaring the negative numbers gives you symmetric answers to the positive numbers, so you’re only doing half the arithmetic (plus it highlights why you get symmetric results when squaring in general).

[@nikoramdorf2886]

the function f(x)=x+1232/x which is the continnual description of the sum of a factorpair of 1232 has a minimum at sqrt(1232) thus the smallest possible sum is 2*sqrt(1232)>2*sqrt(1024)=2*sqrt(32^2)=64 as 54 is smaller than 64 it is therefore to small to be a factorsum of 1232

[@petersievert6830]

I looked at it the other way round equivalently, that the highest product of positive numbers adding up to 54 is 27² < 1232. ;-)

[@quickyummy8120]

@@petersievert6830 I found another amazing Olympiad problem here kzbin.info/www/bejne/oqGqqqt3hpughMk

[@varunvarun4446]

@@quickyummy8120 good question dude

[@jimschneider799]

@5:08 - you can also conclude that n is even by considering the residues of 3^n, modulo 4, which are 1 or 3. Since 3 is a quadratic nonresidue, modulo 4 (and 1232 is congruent to zero, modulo 4), only the cases where 3^n is congruent to 1, modulo 4, (that is, n is even) are possible.

[@ZedaZ80]

I'm still trying to figure out the logic about this in the video (I think I'm about to get it though :P) Your comment worked much better for me just because I'm more familiar with mod 4 fun

[@PunmasterSTP]

Another testament that there is a very educational, professional and wonderful side of KZbin. Thanks for sharing all of your knowledge!

[@VaradMahashabde]

"that's something you learn in elementary number theory class" I asked the prof teaching it in my place to enroll, and he said no

[@KaptenKetchup]

Fun fact: This problem originally came from the Bologna Math Olympiad in 1232.

[@all462]

That seems way before our time

[@taizbin4929]

1932 or 1232?? i think student in 1232 cant do this equation

[@KaptenKetchup]

@@taizbin4929 Nope. Pretty sure it was 1232 ;)

[@taizbin4929]

@@KaptenKetchup woww.. amazing fact

[@quickyummy8120]

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[@goodplacetostop2973]

10:38

[@maxwellsequation4887]

So. Quick.

[@beautyofmath6821]

Great

[@somasahu1234]

Why u went upto k=5 and not beyond that ?

[@quickyummy8120]

@@maxwellsequation4887 I found another amazing Olympiad problem here kzbin.info/www/bejne/oqGqqqt3hpughMk

[@quickyummy8120]

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[@TrimutiusToo]

The smallest is something like 28+44 but check whether 54 can have solutions even if it is clearly too small as well

[@dujas2]

The largest possible product of 2 numbers that sum to 54 would be 27x27, which is less than 900.

[@otakurocklee]

Exactly. 54 cannot give a solution.

[@quickyummy8120]

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[@TheQEDRoom]

Whenever I see a product of the form (a+b)(a-b), I try to find the GCF so I will be left with relatively prime factors. In this case, notice that the GCF of (3^k+m) and (3^k-m) is 2 (since the RHS is even). And so we can reduce this to ((3^k-m)/2)*((3^k+m)/2)=(2^2)*7*11. Since we know they the factors in the LHS are relatively prime, we can reduce the number of cases to 3.

[@GiornoYoshikage]

Another way to show that "n" can't be odd is to reduce mod 8 - we'll wave m² Ξ 3 (8) for odd "n", but only 0,1,4 are squares mod 8. I agree that using mod 4 is simpler

[@absolutezero9874]

Hi m² + 1232 = 3ⁿ 3ⁿ is odd for all n ∈ ℕ 1232 is even Hence, m² is odd m is odd Hence, Let m = 2p + 1, where p ∈ ℕ Hence, (2p + 1)² + 1232 = 3ⁿ 4p² + 4p + 1 + 1232 = 3ⁿ 4p² + 4p + 1233 = 3ⁿ 4p² + 4p = 3ⁿ - 1233 4p(p + 1) = 3ⁿ - 1233 p and (p + 1) are consecutive numbers Hence, p(p + 1) is a multiple of 2 Hence, p(p + 1) ≡ 0 mod 2 4p(p + 1) ≡ 0 mod (2 x 4) 4p(p + 1) ≡ 0 mod 8 Hence, 4p(p + 1) is a multiple of 8 Therefore, (3ⁿ - 1233) is a multiple of 8 (3ⁿ - 1233) ≡ 0 mod 8 1233/8 = 154 r 1 Hence, 1233 ≡ 1 mod 8 Hence, 3ⁿ ≡ 1 mod 8 8 + 1 = 9 Hence, 9 ≡ 1 mod 8 3² ≡ 1 mod 8 Hence, (3²ᵏ) ≡ 1 mod 8 , where k ∈ ℕ Let’s try 3⁶ Hence, k = 6/2 = 3 3⁶ = (3²)³ = 9³ = 729 3(729) = 2187 Hence, 3(729) > 1233 Hence, 3⁷ > 1233 Therefore, 3⁸ > 1233 Hence, 3⁸ is the smallest number that is greater than 1233 where n is even All Hence, Sub. n = 8 into given equation: m² + 1232 = 3⁸ = 3 x 2187 = 6561 m² = 6561 - 1232 = 5329 m = √5329. or. m = -√5329 = 73 = -73 (rej. as m ∈ ℕ) But when I sub. in n = 10, m² is not a perfect square Why?

[@klausg1843]

Taking modulo 4 on both side shows that n is even, say = 2k, since m^2 =0 or 1 mod 4. Then factorize to get (3^k -m)(3^k+m)=1232= 2^4•7•11. Since the sum of the factors is = 2•3^k, only one of the factors contain the factor 2 exactly once. The other factor will get the factor 8. Now its straight foreward to enumerate the 4 possibilities for the two factors: (2 , 8•7•11), (2•7, 8•11), (2•11, 7•8) and (8, 2•7•8). Only the last one is a possible solution, giving n=2k=8 and m=73, which satisfy the given equation.

[@echandler]

Nice problem and solution. My arguments go as follows: m^2 + 1232 = 3^n for m,n in N 3^n is odd and 1232 is even, thus m must be odd the only quadratic resudues mod 4 are 0 and 1, residues of powers of 3 are 3 (n odd) and 1 (n even). Thus n is even. n = 2k, k in N 3^(2k) - m^2 = 1232 = (3^k+m)(3^k-m) = pq where p,q are in N and p>q 3^k and m are both odd thus p and q are both even. p-q = 2m thus gcd(p,q) | 2m where m is odd, thus 2 is the highest power of 2 in d so the powers of 2 must be split 8,2 or 2,8 between p and q. This leaves 7*11 to divide among p and q however p > q. This leaves only 4 possible factor pairs for (p,q) (56,22) (88,14) (154,8) and (616,2) only (154,8) sums to 2*3^k for k=4, thus m=73, n=8

[@anshumanagrawal346]

Nice solution

[@brinzanalexandru2150]

The easiest way to get that n is even is mod8,by contradiction assume that n is odd so n has the form 2k+1 where k>0 so 3^n =3*9^k now because 9 is 1mod8 then 9^k is 1 mod8 so the whole thing will be 3mod8,and because 1232 is 0mod8 it means that m² is 3mod8 which is impossible because it can give only 0,1,4mod8.

[@manucitomx]

Thank you, professor!

[@mirjavadmiremarati4972]

There is another way to show "n" is not odd : If n is odd then 3^n ends to "3" or "7" then m^2 must end to "5" or "1". but we know if m^2 ends to "5" it ends to "25" or if m^2 ends to "1" it ends to "x1". (x is even.) Therefore, m^2+1232 ends to "y3" or "z7". (y and z are odd.) But, in other hand, we can show 3^n ends to "u3" or "v7". (u and v are even.) and there is a contradiction! thus, n is not an odd number.

[@aadarshsavaikar4337]

Can you please solve this question: ABC-CBA=CAB, A, B, C≠0 & A>B>C

[@martinlabrousse8865]

@michael, you have forgotten that 3^6 = 729 = 1 mod 7 Not sure that it can give solutions for 3^ (7i+6) but you still have to prove that n (of n^2) is peer

[@Ranoake]

I did not follow what he meant by the only time you get perfect squares is when the exponent is even (4:54). 2 is not a perfect square?

[@isaacdeutsch2538]

Perfect squares mod 7. You'll notice that in the first table, the only numbers in the bottom row are 0, 1, 2, and 4. That means that any perfect square has a quadratic residue of 0, 1, 2, or 4 mod 7, that is, if you divide any perfect square by 7, you'll get a remainder of 0, 1, 2, or 4. So because 2 is a quadratic residue mod 7, it is equivalent to a perfect square, if you're dividing everything by 7 and taking the remainder. Essentially his argument is that if you mod 7 both sides of the equation-that is divide both sides by seven and take their remainders--they must be equal, because they were equal at the beginning. So since a perfect square mod 7 can be 2, so can 3^n. Does that help?

[@Ranoake]

@@isaacdeutsch2538 OK, I get it now. I am too old for this stuff haha. Thanks!

[@georgesadler7830]

Professor Penn, thank you for a cool problem in the Number Theory archives.

[@rickenbackerlover7386]

I was watching your video about the divergence of the sum of the reciprocal of primes and wondered about the sum of the reciprocal of primes squared, it probably converges, but does it have a closed form? Can you show it? Thanks!

[@megauser8512]

In fact, it definitely converges, since it is > 0, and it is < the sum of the reciprocals of all positive integers squared, which already converges to pi^2 / 6.

[@theevilmathematician]

If we rearrange, we get 3ⁿ - m² = 1232. There are 2 cases for n. If n is odd, we would get a contradiction, meaning n is even. If n is even, then let n=2k, for a natural number k. Then 3^(2k) - m² = 1232. By the Difference of Squares, we get (3^k + m²)(3^k- m²)= 1232. The prime factorization of 1232 is (2⁴)(7)(11). Then, the equation becomes (3^k + m²)(3^k- m²)= (2⁴)(7)(11). However, (3^k + m²)(3^k- m²) are factors of 1232 and thus, must be in the form 2+ 3^k when we take the sum of the factors. The only case that works is when 1232= 8 * 154, giving 162 when adding 8+154. Therefore, this leaves a system of equations: 3^k- m=8 and 3^k + m=154, meaning k=4 and m=43 and n=8.

[@mrphlip]

I did the first half similarly, but with different moduli: 3^n is odd, and 1232 is even, so m^2 must be odd, so m must be odd This means that m^2 = 1 (mod 4), and 1232 = 0 (mod 4), so 3^n = 1 (mod 4) The powers of 3 alternate between 1 and 3, mod 4 (since it's the powers of -1), so this implies that n is even. Then proceeding as in the video, we need factors of 1232 whose average is 3^a, and whose difference is 2m. Since the difference is even, both factors must have the same parity... so both factors are even. But since m is odd, the two factors cannot both be multiples of 4 - one is, and the other is not. So one is a multiple of 2, and the other a multiple of 8. This gives us just 4 pairs of factors we have to consider: 2*616, 14*88, 22*56, 154*8. Checking each one by hand shows that 154*8 is the only one where the average of the factors is a power of 3. So m = 73, n = 8. I'm confused by the "Multiple solutions!" on the video thumbnail as there only appears to be one solution, unless I'm missing something?

[@anshumanagrawal346]

I think that's a mistake/attempt to confuse you into thinking 54 is possible

[@2011mrchepe]

I thought of this problem by observing that 3^n is always odd. This allows us to look for some square that is odd so that m^2+ 1232 is odd. Now if n

[@2011mrchepe]

1232*

[@2011mrchepe]

Other typo meant to say m^2 = 5329

[@quickyummy8120]

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[@quickyummy8120]

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[@harshparashar9210]

Great video 👍 Keep it up

[@charlesglidden557]

Why does 3**n need to be a perfect square? M**2 is a perfect square but not necessarily m**2+ 1232

[@alihamchou9278]

Good morning Michael , which is the best math book for year 11 and 12 Thank you

[@neilgerace355]

4:50 can someone please explain how 2 is a perfect square in this sense?

[@isaacdeutsch2538]

Perfect squares mod 7. You'll notice that in the first table, the only numbers in the bottom row are 0, 1, 2, and 4. That means that any perfect square has a quadratic residue of 0, 1, 2, or 4 mod 7, that is, if you divide any perfect square by 7, you'll get a remainder of 0, 1, 2, or 4. So because 2 is a quadratic residue mod 7, it is equivalent to a perfect square, if you're dividing everything by 7 and taking the remainder. Essentially his argument is that if you mod 7 both sides of the equation-that is divide both sides by seven and take their remainders--they must be equal, because they were equal at the beginning. So since a perfect square mod 7 can be 2, so can 3^n. Does that help?

[@casusincorrabilis1584]

He described it wrongly. He should have said "Lets look when the left and right side can be kongruent to mod 7 by checking the results if we make a table of possible inputs" and because of the cyclical characteristic of the mod function it is obvious that only even y can ever lead to a solution. I wonder how the first people come to that Idea, because it's really,really tricky thing and only gives you that single bit of more information needed to solve the puzzle. It wouldn't come to my mind normally to even try this to get an easier equasion. That's the magic of Numbers Theory.

[@isaacdeutsch2538]

@@casusincorrabilis1584 I don't know that he described it incorrectly, I think he was just not as clear as he could have been. I think he assumes the viewers are familiar with modular arithmetic.

[@quickyummy8120]

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[@quickyummy8120]

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[@doctorkrusher3236]

I would have subtracted 1233 from both sides.

[@kartikdas8611]

You are doing an amazing thing,love from Bangladesh.

[@kemalkayraergin5655]

that was pretty sweet

[@somasahu1234]

Why u went upto k=5 not beyond that ?

[@lexyeevee]

after that is 1458, and the greatest possible sum of factors of 1232 is 1232 + 1 = 1233

[@arnabroy2247]

Awesome 😍

[@otakurocklee]

How are you getting 54 gives a solution?

[@shreyamjha3058]

Mod 4 implies 3^n=m² mod 4= n is even as m² can't be 3 mod4 ,bunch of tedious calculations saved ,reached 5:07 in a single statement 😁 ,and also m is odd

[@TJStellmach]

You seem to have been trying to say something.

[@megauser8512]

That makes sense.

[@clintongryke6887]

George Sassoon would love these videos.

[@drsonaligupta75]

Just take Mod 4

[@johannesh7610]

Yes, you can do the same thing as he did with 7,but are much quicker

[@holyshit922]

I guessed only one solution (73,8)

[@mithutamang3888]

Yes, 54 is an also a solution!!! 😁😁👍👍

[@memomariya2101]

？？？？？really ? ?

[@mithutamang3888]

@@memomariya2101 OK

[@megauser8512]

Ok, you have to be joking, because 2 * 3^k = 54 implies that k must = 3, so the equation becomes 3^(2k) - m^2 = 1232, which becomes 3^(2*3) - m^2 = 1232, and in turn we have 3^6 - m^2 = 729 - m^2 = 1232, so -m^2 = 503, so m^2 would have to be -503, but that makes m NOT a whole number!

[@mithutamang3888]

YES 54 IS A SOLUTION BUT, ALSO WHOLE NUMBER!!! 😁😁👍👍

[@mithutamang3888]

54 works!!! 😁😁👍👍