When is this a perfect square?

[@pratikmaity4315]

This is from IOQM(Indian Olympiad Qualifier in Mathematics) of 2021

[@daivikdev7575]

Yaa

[@MohitKumar-ds6ok]

its was from 2020, 2021 exam will be in August

[@areein4735]

I appeared for it..wasnt able to solve tho

[@pratikmaity4315]

@@areein4735 no problem at all. The only thing that matters is what the problem taught you.

[@sakethram538]

yes and i was able to solve it

Interesting. Both Michael and "blackpenredpen" make the same mistake: 2^8=128. But 2^8=256. Nevermind, good as usual. Thanks Michael.

[@igormerling]

2^8 is 256, and not 128 (which doesn't invalidate your proof, but anyway)

[@michaelempeigne3519]

It invalidates the proof if numbers are wrong practically.

[@petersievert6830]

@@michaelempeigne3519 Then let's put it that way: the proof would work just as well with the correct number in place, so no big harm?

[@MichaelPennMath]

Obviously 2^8=256...lol

[@sergiokorochinsky49]

byte me

[@vietphan3767]

Nearly have a heart attack there

[@sporksto4372]

r/beatmetoit

[@tashkiira7838]

I was about to say.. :D

[@anshum1675]

@@sergiokorochinsky49 underrated joke lol

[@leontaylor3721]

Hi Michael, i just wanted to say i love your content, and i appreciate the passion and effort you put into your videos. Keep the energy flowing my friend

[@asklar]

There's something deeply unsatisfying *to me* about manually checking different cases, I wonder if others feel that way too, and whether you have any insights into why that might be. Have we not developed methods that can give closed form solutions that don't involve brute force? Or is there something more fundamental about the structure of the problem and about number theory in general?

[@TechToppers]

Brute force is common in Number Theory. As for checking cases, yeah it looks unintuituve to check but, you have to notice absolute value sign and plug in values to understand how the function behaves over n in integers. That is a common thing to do in any branch of Math ig.

[@cobalt3142]

Another thing to add: the four color theorem's best current proof still involved a computer checking 938 different cases (I think that's the right number?), so sometimes casework is the best we can do

[@SlidellRobotics]

On "I'll let you check this isn't a perfect square" just before 8:00, pretty trivial. 64 + 5⁶ - 65 = 125² -1. The only squares that differ by one are zero and one.

[@danielleza908]

nice!

[@IanXMiller]

Find all numbers which are both square and triangular. Or phrased a bit differently find the recurrence relationship between consecutive terms (or closed form if you prefer) in OEIS A001110 (all integers which are both square and triangular). The formula is quite unusual.

[@thenateman27]

This function is very similar to the Fermi-Dirac distribution function we use in physics. The function looks like f(E) = exp(-x)/(1 + exp(-x) ) where x = (E - E_f)/kT. As we take T to absolute zero, or the limit as T goes to 0, the function turns into a step function around the Fermi Energy E_f.

[@GOATvoldemort]

This question was in Mathematics olympiad of India (IOQM) 2021

[@Wurfenkopf]

This is brilliant!!! One question: when you receive viewer suggested problems, like this one, do they give you the solution or do you find it yourself? Thank you!

[@AlephThree]

Nice, I got it up to constructing the inequality. A bit more practice needed.

[@boredgamesph4872]

@7:51 why you use m?

[@wlong8901]

8:01 64+5^6-65 = 5^6 -1 = (5^3)^2-1 which is not a perfect square. No need to do out calculation.

[@chrstfer2452]

Youre such a funny dude, mr penn. Ive quite enjoyed binging your catalogue recently, and its huge so i dont anticipate running out anytime soon. Slowing working through your differential forms playlist with pen and paper beside me too, as prep for further diving into tensors. Thanks!

[@gamedepths4792]

OMG This is from the ioqm exam recently held in India which I gave as well!! Unfortunately, this problem really stumped me as i didn't get any time to do it

[@36sufchan]

this is from the IOQM 2021! holy damn, never thought I'd see a problem in an exam before I see it on KZbin

[@shivansh668]

A special thanks Michael ! I 've suggested you this Problem.

[@goodplacetostop2973]

11:16

[@cyrenux]

2 weeks early

[@oliverbrooker882]

Bro how tf you comment 2 weeks before release

[@goodplacetostop2973]

@@oliverbrooker882 I’m a time wizard

[@vedants.vispute77]

@@goodplacetostop2973 no u r a just good place to stop

[@wasitahmid749]

Is there anyone who looks this section only for homework??? Give some (interesting) math problems man!!!

[@leif1075]

Isnt there a more intuitive way to solve than using the mod stuff ?? Especially the part for the cases where n is greater than 3?

[@contestmath6257]

You do not have to calculate n=6 by hand because 64+5^6-65 = (5^3+1)(5^3-1) which can not be a suquare because the factors only share prime divisor 2.

[@vinc17fr]

Similarly, I did 2^6 + 5^6 − 65 = (5^3)^2 − 1, which cannot be a square as the only two squares that are consecutive integers are 0 and 1.

[@contestmath6257]

@@vinc17fr Oh yea thats even faster. Nice!

[@ZedaZ80]

Here is a fun problem: I noticed at work one day that 343-243=100 ==> 7^3-3^5 = (7+3)^(5-3) What are the (other) solutions to a^c-b^d=(a+b)^(d-c) where a, b, c and d are natural numbers? It has been a while, but there are two other solutions that I know of (found by a professor friend). It has been a while, but I think this was the problem I was working on that I wrote a program for that ruled out all other solutions with numbers less than 10^1500, but that might be another problem.

[@ZedaZ80]

It's all coming back to me now :P There are trivial solutions (where the powers are equal, for example). I found the program that I was thinking of, but of course I left no documentation about the problem it was for (just a vague, "this will find the solutions" and notes about upper and lower bounds).

[@leif1075]

At 3:40, how is that whole expression congruent to 2 to the n mod 5..thst doest make sense..if you divided 2 to the n by 5, the remainder wont be 2 to the n..Case in point when n equals 3, the whole expression equals 68 which is 3 mod 5, NOT 2 raised to the 3 mod 5...

[@ricardocavalcanti3343]

2^3 (mod 5) = 8 (mod 5) = 3 (mod 5).

[@Famoke]

i have one problem like this in middle school: Find all positive integers x, y, and k such that 2^x + 5^y = k^2

[@aa-lr1jk]

Here goes a nice (and relativelly hard) problem: let f: R^3->R^3, such that |p-q|=1 implies |f(p)-f(q)|=1. Show that f is a isometry (i.e. |f(p)-f(q)|=|p-q|, for all p and q).

[@СергейСергеенко-щ8д]

To speak about isometry we need to define metric space. What are distances in your problem?

[@aa-lr1jk]

@@СергейСергеенко-щ8д The usual euclidian metric: |x| = (x_1^2+x_2^2+....+x_n^2)^(1/2)

[@iamadooddood4331]

5^n - 65 is a multiple of 10 so if we use mod 10 instead of mod 5 it'll be easier to see that only even values of n can result in squares. It also follows that all possible squares here are squares of even numbers. For m = 3 it's easy to see that it doesn't result in a perfect square because 5^6 + 2^6 - 65 = 5^6 - 1; here we get (N-1)² < N²-1 < N² where N = 125. Also lol at repeating the usage of m for that m². Finally 2^8 is 256, not 128.

[@maxlagus9042]

Hey, I really love your videos! Can you please show some graph problem? I think it is very interesting and hard term in olym math

[@jamescollis7650]

at beginning, n=2 or 4 mod 5 does not imply n is even. Consider the initial equation mod 4 to deduce n is 0 or 2 mod 4 and hence even

[@failsmichael2542]

He means that by Fermat, 2^4 = 1 mod 5 so 2^n = 2^(n mod 4) mod 5. So to determine 2^n mod 5, it suffices to consider n mod 4. And he showed that 2^n is a square mod 5 only when 2^n = 1 or 2^n = 4 mod 5, which happens for n = 0 or 2 mod 4, i.e n is even.

[@basmamjouel2893]

@@failsmichael2542 why does it imply that 2^n=2^(nmod4). Mod5

[@stvp68]

There was an add literally 2 seconds before the end-between “and that’s a” and “good place to stop”. Yeesh!

[@jbtechcon7434]

I have a question, not quite what you usually do on this channel, but it might make for an interesting discussion. What would happen if we woke up tomorrow morning to find that someone had published a clear counter-example to the Reimann Hypothesis? It would have to be a very large number, since it's already been proven that no counter-examples exist with a small magnitude... But what would it MEAN? How would maths change? What work would all the mathematicians need to redo or rethink with this counter-example in mind? I've asked this question on StackExchange, but all it got was nominated for deletion.

[@dionisis1917]

Find all natural number x,y,n such that: 2^x+5^y-65=n^2

[@Fun_maths]

2^8=256 not 128 but that doesnt change anything

[@davidcroft95]

Good demonstration! Just a little question: when we did theory number, my math professor usually (or rather always) uses module 8 to understand if a number were a perfect square (since module 8 every perfect square is either equal to 0, 1 or 4). How this could have changed your demonstartion?

[@levkrainov]

The tool you use has to capture the essence of the problem in hand. Here mod 5 shows us that n has to be even and then we have (5^m)^2 + 4^m - 65. Since 5^m grows so much faster than 4^m this addition can never cover the gap between (5^m)^2 and the next perfect square. This allows us to eliminate all high enough numbers n. To modify this to use mod 8 would probably imply proving that n=2m using mod 8, leaving second part of the proof intact.

[@davidcroft95]

@@levkrainov oh, okay! Thank you for the answer!

[@reeeeeplease1178]

You try to choose the mod in such a way that as many parts of the equation vanish This makes handling the rest easier Mod 8 would have turned the 65 into a 1 and the rest would have stayed the same which doesnt really help here Mod 5 canceles the 5^n and the -65 completely, leaving us with only one term

[@davidcroft95]

@@reeeeeplease1178 yeah, that makes sense. Thank you for your answer!

[@TechToppers]

I had the same solution. Not sure how modulo 10 works...

[@prithujsarkar2010]

Good memories while solving this problem using some engineering lol

[@deepjyoti5610]

Tu engineer kb se ban gya

[@subhamshreyash7906]

Op

[@swastikadeb8605]

Lol

[@TechToppers]

You can't *solve* it by engineering.😂😝

[@leif1075]

What engineering?

[@leif1075]

How does he apply Fermat's little theorem here if the exponent is not 5 minus 1 it is n not n minus 5 or 5 minus n which fermats theorem says the number is raised to p minus 1 when p is the prime aka 5 in this case.. or just raised to p if you multiply both sides by the base ..that's not what we have here...

[@ricardocavalcanti3343]

You can write any integer n as n = 4m + k, where m is an integer and k = 0, 1, 2 or 3. Therefore, 2^n = 2^{4m+k} = (2^4)^m x 2^k = 2^k (mod 5), since 2^4 = 1 (mod 5) by Fermat's little theorem. That's why he says that one needs only to look at values of n between 1 and 4 (or between 0 and 3, as 2^4 = 2^0 (mod 5)).

[@leif1075]

@@ricardocavalcanti3343 yea butbhebsays the whole expression 2 to tbe n plus 5n minus 65 equals 2n mod 5..not just the 2 part..the way you said it is clearer..not the way he said it in the video

[@nirorgad]

Thank you for all these videos reminding me of my number theory classes in university of Tel-Aviv, Israel :)

[@zrksyd]

You can prove 64 + 5^6 - 65 is not a perfect square without calculating by hand. It becomes 5^6 - 1 which is one less than a perfect square that isn’t 2 since 5^6 is a perfect square.

[@aleksapupovac]

Isn't 2⁸=256?

[@aleksapupovac]

Seems it doesn't affect the solution, but I still wanted to point out the mistake.

[@tahirimathscienceonlinetea4273]

It's a great proof

[@hendrixgryspeerdt2085]

pretty crazy at 2:06

[@michaelf.7050]

Doesn't this only prove that there cannot be any more solutions between a power of five and the integer above? Or am I missing something?

[@rtyk1]

He proved that for m>=4, the quantity 2^n+5^n-65 falls between 2 consecutive squares. The deduction is then that this quantity cannot be a square, otherwise we contradict the fact that the squares it falls between are consecutive. So there are no solutions for m>=4

[@void7366]

Hey @Michael Penn, i have this pretty interesting problem for you to look.at if you want, ive been trying to solve it for a while but to no avail, the problem goes as follows: n is a natural number Proof there does not exist a natural number k such that : 3n²+3n+7=k³ Thank you in advance! (Edit) ive arrived to a certain result after trying this problem out for the second time but i'm not so sure it's true, if anyone can help, plz do, i did as follows: Suppose BWOC 3n²+3n+7=k³ is true for some n;k€N 3n²+3n+7=k³ n²+(n²+2n+1)+(n²+4n+4)+2=k³+3n n²+(n+1)²+(n+2)²=K³+3n-2 -----adding the first (n-1)² terms to both sides ---- 1+2²+....(n-1)²+n²...(n+2)²=1+2²+....(n-1)²+k³+3n-2 There's a nice formula for the sum of the first m² terms, applying it.to both sides gives: n(n+3)(2n+1)/6=n²(2n+1)/6 +k³+3n-2 We take everything to the left hand side except for k³ and after simplifying we arrive at (2n²-5n-4)/3=k³ My observation is that since all we did was add and subtract natural numbers from both sides all along, still have the same k³ term that we started the argument with, furthermore k³ in the argument also satisfied the equation 3n²+3n+7=k³ So 3n²+3n+7 must equal ( 2n²-5n-4)/3 and which has no solutions in the set of natural numbers, which then tells.us that our assumption is false.. I hope someone reads this and corrects me if im wrong i will be so grateful

[@ricardocavalcanti3343]

The sum of the first n squares is n(n+1)(2n+1)/6, therefore the sum of the first n+2 squares is (n+2)(n+3)(2n+5)/6, and the sum of the first n-1 squares is (n-1)n(2n-1)/6. Here's my solution: 3n^2+3n+7 = 1 (mod 3) => k^3 = 1 (mod 3) => k = 1 (mod 3) => k = 3m+1 => 3n^2+3n+7 = (3m+1)^3 = 27m^3+27m^2+9m+1 => 3n^2+3n+6 = 9m(3m^2+3m+1) => n^2+n+2 = 3m(3m^2+3m+1) => n^2+n+2 = 0 (mod 3). (*) But (*) has no integer solution, since (1) n = 0 (mod 3) => n^2+n+2 = 2 (mod 3), (2) n = 1 (mod 3) => n^2+n+2 = 4 (mod 3) = 1 (mod 3), and (3) n = 2 (mod 3) => n^2+n+2 = 8 (mod 3) = 2 (mod 3). Therefore, the original equation also has no integer solution.

[@RexxSchneider]

When you're dealing with cubes, the go-to base for modulo arithmetic is 9 because it has only 3 residues: {0, 1, 8}. Note that 3n^2 + 3n + 7 = 3n(n+1) + 7 and 3n(n+1) mod 9 can only be 0 or 6 for n>0. You only have to calculate the first three, after which it repeats. So 3n^2 + 3n + 7 ≡ 7 or 4 mod 9 and neither of those can be a perfect cube mod 9. And that's the place to stop.

[@void7366]

@@RexxSchneider thank you guys so much

[@sharpmind2869]

Nice problem from IOQM 2020

[@Lionroarr]

Really nice

[@mrosskne]

brain of a maths god, voice of a radio jockey

[@mrmanning6098]

New variant: When is 2^n + 5^n -65 square-free?

[@anshum1675]

Whenever it's not a perfect square

[@mrmanning6098]

@@anshum1675 ​ Is that true? Square-free means that it has no perfect square factors, or that each of the powers in its prime factorization are no more than 1

[@anshum1675]

@@mrmanning6098 Oh I didn't know lol. I thought square free meant not a perfect square. So I thought your question was weird

[@RedRad1990]

@@anshum1675 Admit it. You thought it's a joke, a troll or someone stupid 😜

[@anshum1675]

@@RedRad1990 yeah I thought it was a joke lol

[@michaelempeigne3519]

2^8 = 256

[@_judge_me_not]

For how many n € N , n⁴+11n²+1 is a perfect square?

[@ricardocavalcanti3343]

One, if 0 is included in N.

[@_judge_me_not]

Thanks 😀

[@aleksmich8928]

Hi, Michael. Greetings from Lithuania. Perhaps you would be interested in a problem from Lithuanian national math contest. The problem is to find the remainder of (1^965 + 2^965 + 3^965 + ... + 2018^965) modulo 2021. This math contest is organized by Kaunas University of Technology, which I currently work at (teaching and research). The contest consist of two rounds. In the first round we give 10 problems and ask for the answer only (no solution necessary) whereas in the second round we give two problems and ask for a complete solution. The suggested problem was given this year to 11th year students (17-18 years old) in the second round.

[@goodplacetostop2973]

You can submit your problem there : forms.gle/XZyx9N3WJzdRCjM49 so your suggestion won't get lost in the comments

[@aleksmich8928]

Thanks!

[@suniltshegaonkar7809]

excellent!

[@fansuli427]

very good video thanks

[@conovan5081]

Good stuff

[@vinc17fr]

I first considered the case n even as this was obvious; similar method already seen in kzbin.info/www/bejne/d6Ocm2WGiriofqM And for the case n odd, I considered mod 3, which is simpler.

[@mohammadshaanmdsalamsaikh6010]

SAME: IOQM 2021 INDIA ... WONDERFUL... !!

[@pratikmaity4315]

Hi Michael here is a nice Diophantine Equation you can make a video on. (Japan 2020 junior finals problem 3) Find all tuples of positive integers (a,b,c) such that Lcm(a,b,c)=(ab+bc+ca)/4. Thanks and I am enjoying your videos and pedagogy a lot.

[@rockinroggenrola7277]

You can suggest a problem on the google form now.

[@pratikmaity4315]

@@rockinroggenrola7277 Thanks.

[@leoleo3783]

Thank you for great videos, I hope more videos will be published!!! In my country, It is supposed that N includes 0 and N* doesn’t, so I think 0 should be added in the first case although it is not satisfied.

[@SONUKUMAR-vr2jg]

Thanks for your nice solution 🇮🇳

[@saptashwabaisya4202]

IOQM 2021 Question

[@АлександрКотов-н4к]

Michael's student(girl): I know you're married, but you're so hot. Michael: and that's a good place to stop.

[@Fun_maths]

@ehud kotegaro I'm with you on this one, this is a math channel and not quite the appropriate place to comment that

[@trishanmondal7813]

IOQM 2021 problem from india . :) I did it during exam and got correct

[@rajvardhansinghsisodiya1095]

My bad I didn't able to solve in exam

[@hgtmonster4857]

This is ioqm problem

[@klementhajrullaj1222]

Oh my God, your hairs again! 😲😲😲

[@דףאחדעלמתמטיקה]

אם אתם דוברי עברית, אתם כנראה תהנו מהערוץ שלי 'דף אחד על מתמטיקה'. בואו לבדוק :)

[@oliverbrooker882]

I think you may have used 2 different m's in your working I got a lil confused for a second 👉🏻👈🏻

[@mariomestre7490]

Genial, crak!!

[@debayuchakraborti1963]

It was a tough question for me to solve during the exam...i tried it for like 5 mins and left cuz the time pressure was huge!!!!!

[@The_Math_Enthusiast]

I did not solve it on the test(LOL) came back home and solved it out.

[@MANISHGUPTA-oe3in]

In which class you are ? @Debayu Chakraborti

[@random-td7tf]

Haha lol same here but I knew it would have small answers. Found 2,4 work so 2+4=6. At the end of the exam, proved the values were correct

[@thayanithirk1784]

From ioqm 2021 india

[@mukulsinghal3164]

Yaa correct

[@shubhamjha3064]

Legends know that this was suggested by an Indian teacher or student as it came in IOQM 2021....... Question number 13

[@ronludmer126]

8^2=256