A number theory proof

Find integer solutions a^2+b^2=4c+3 , a number theory proof or disproof.

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[johnhumberstone]

My guess is that the people you teach face to face, have no idea how lucky they are.

[RasperHelpdesk]

At the point you showed a2 + b2 = 4(k2 + l2 + l) + 1 My thoughts went this way:
a2 + b2 = 4c + 3
a2 + b2 = 4(k2 + l2 + l) + 1
4(k2 + l2 + l) + 1 = 4c + 3
4(k2 + l2 + l) = 4c + 2
k2 + l2 + l = c + 1/2

Since k and l are integers, (k2 + l2 + l) is an integer, but since c is an integer, c + 1/2 CAN'T be an integer, so there is no solution.

[styzon]

For any integer x, x=0, 1, 2, 3 modulo 4. If x=0 or 2 (mod 4), the x^2=0 (mod 4). If x= 1 or 3 (mod 4), the x^2=1 (mod 4), so the square of any integer is 0 or 1 modulo 4. Thus a^2, b^2 is either 0 or 1 module 4. So their sum is at most 2 module 4. If a^2+b^2=4c+3, then a^2+b^2=3 (mod 4), which is not possible. Thus we have no integer solutions.

[emmeeemm]

That's fantastic! I did it basically the same way, except I did more algebra and worked it down to

2(k^2+l^2+l) = 2c+1

And then, I noticed that my left side was even and my right side was odd, so I concluded that there were no solutions. Going straight for the mod-4 observation was pretty slick!

[aniruddhvasishta]

Just saying, any integer squares are either 1 or 0 mod 4, so this proof is easy if u know that.

[waatup]

I used modular arithmetic. When an integer n ≡ 0 or 2 (mod 4), n^2 ≡ mod 4; when n ≡ 1 or 3 (mod 4), n^2 ≡ 1 (mod 4). Then the sum of two integer squared can be congruent to mod 4, 1 (mod 4), or 2 (mod 4), but not 3 (mod 4). Therefore, a^2 + b^2 can't be equal to 4c+3 if a, b, c are all integers. Mod made it pretty simple and it took me only two minutes lol

[alexpagnetti]

"a is 2k, not like the 2k18 for the basketball games, but 2k" 😂

[johnsmith]

once he wrote the 4(k^2+L^2+l) +1 i was like It looks so obvious once he points it out but at the start I'm just wondering what the heck to do. Love the video.

[nozack]

Or directly, since (k^2 + l^2 + 1) is an integer, denote it as d.

Then
4d^2 + 1 = 4c^2 +3
4d^2 = 4c^2 +2
d^2 = c^2 + 1/2

Since d, c and their squares are all integers, they cannot differ by 1/2.

[OonHan]

Wait... so if it was 4c+5 instead, there would be a solution?? (Or at least 4c+(4n+1))

[leoitshere]

Suppose integer solutions exist, then take the congruence mod 4

a^2 + b^2 = 3 mod 4

But then mod 4 squares can only equal 1 or 0 (you can prove this by computation). That means that a^2 + b^2 can only be 0, 1 or 2 mod 4. Not 3. A contradiction. Thus, no solutions exist.

[baconman]

Interestingly, there is a solution for the equation:
a²-b² = 4c+3
In the case where a=2k and b=2l+1, a²-b² becomes:
4(k²-l²-l)-1 which we can re-write as 4(k²-l²-l)+(3-4)
And rearranging this yields 4(k²-l²-l-1)+3
Therefore letting c = k²-l²-l-1 we find that a²-b²=4c+3

[Jonathan_Jamps]

I highly recommend your math videos to my friends. Yeah!

[billprovince]

Different approach: consider both sides of the equation mod 4, ie. (a^2 + b^2) mod 4 = (4c + 3) mod 4.
Note that for any x, it can be decomposed as 4y + z, where z in {0, 1, 2, 3}, and therefore x^2 mod 4 = (16y^2 + 8yz + z^2) mod 4. However, the r.h.s reduces to just z^2 mod 4. Since z in {0, 1, 2, 3}, z^2 in {0, 1}.
Therefore, (a^2 + b^2) mod 4 is in {0, 1, 2}.
However, (4c + 3) mod 4 = 3.
From here, since 3 is not in {0, 1, 2}, we can conclude that there is no solution.
One key trick I find for integer equations is to use modular arithmetic to eliminate variables. In this case, I used mod 4, because I knew that it would eliminate the variable c.

[googleuser]

Simply divide both sides by 4 and you will get different remainders o both sides, therefore no integral solutions are there for a, b, c.

[lwolstanholme]

you're my favourite maths youtuber hands down

[ethanwolbert]

I subtracted 4c + 3 to the other side.
a^2 + -(4c + 3) + b^2 = 0
Once you have that, the inly way for that to factor and have solutions is for -(4c +3) = 2ab. From there (a + b)^2 = 0. a = -b.
Subbing in to the original equation, 2b^2 = 4c + 3
b^2 = 2c + 3/2
The right hand side of the equation is not an integer so therefore there are no solutions since b^2 must itself be an integer.

[megachikendancr]

dang u r so good at math and u r making me love math!! i am in algebra 2 honors, and my teacher was working out that problem you had about sqrt of i and it was EXACTLY what you did so i was like

[sahilhalarnkar]

The sum of two distinct squares gives a number of 4N+1 type. The RHS is a number of 4M+3 type. Therefore they can never be any pair of integers N and M for which the equation is valid.

[oluwagbogoajimoko]

This has been a good start for my number theory journey.