I did it in a more complex way. since RHS is an integer, so LHS must be integer as well, so n has to be even. let n=2k, it follows that k(k+1) = m4+m2-m+1=(m^2+1)m^2 - (m-1) Since RHS has to be product of two consecutive integers, And since m^2(m^2+1) - k(k+1) = m-1; m^2(m^2+1) -m^2(m^2-1) = 2m^2 2m^2>m-1 for any real number m. so (k+1)k>m^2 (m^2-1) is the only solution here. Which means k>=m^2, which means m-1
My method was to set n=2k (following form n(n+2)/4) becouse for this expression every even number will give an integer so then I got the equation k²+k=m⁴+m²-m+1 now move k²+k to rhs and factor in the following way (m²-k)(m²+k+1)=-(m-1) now becouse m,n≥0 it follow that k≥0 but by doing manipulations with that inequalities we get that m²-k≥0 and m²+k+1≥1 but so their product is at least 1 but becouse -(m-1) it would only be possible if m-1≤0 so m≤1 and we. have to check only m=0 and m=1 to get our pair (m,n)=(1,2)
@@tianqilong8366 I should check 2 cases 1)when m²-k≥0 2)when m²-k≤0 Check my second comment in order to see the prove that there are no solutions for the second case
OK, Since RHS is integer, then n must be even. If we write n=2k, then k*(k+1) = m^4+m^2-m+1 But a consecutive product is even, so, in the LHS, m must be odd (So no zero from the begining :))). Now, the expresion m^4+m^2-m+1 must be a product of 2 consecutive numbers. In fact, we don't need n or k next, is anough just that LHS is a product of 2 consecutive numbers. This product, obviously, will be "connected"with m^2-1, m^2, m^2+1. So, we are going to prove that our expresion in m is bigger than a product of 2 consecutive numbers and less than the next product of 2 consecutive numbers. First product: (m^2-1)*m^2 Let's prove (m^2-1)*m^2=0 which is allways true (discriminant is negative). More of that, (m^2-1)*m^2 < m^4+m^2-m+1 (the inequality is strictly) If we take the product of the other 2 consecutive numbers: m^2(m^2+1) Let's prove m^4+m^2-m+1=1. But m is odd, so the inequality is allways true, with equal sign when m is 1. So (m^2-1)*m^2 < m^4+m^2-m+1 -m+1=0 => m=1 Therefore m=1 is the only solution. And then n=2, of course.
I saw this trick some times to find when an expression is a square, it’s very useful to reduce the possible values of some numbers and then make calculuses with easy values. A simple examples are “find when 9n^2+36n+76 is a perfect square” or “solve 16n^2+7n+4=m^2”
Whybeven add 1 after multiplying bith sides by 4..why not just set both sides as a product..n can equal 4 and then n plus 2 equals equals rest ofnthe expression and solve or n equals 2 and n plus 2 equals 2 times all the other terms and so on..
I did it in a more complex way.
since RHS is an integer, so LHS must be integer as well, so n has to be even.
let n=2k, it follows that k(k+1) = m4+m2-m+1=(m^2+1)m^2 - (m-1)
Since RHS has to be product of two consecutive integers,
And since
m^2(m^2+1) - k(k+1) = m-1;
m^2(m^2+1) -m^2(m^2-1) = 2m^2
2m^2>m-1 for any real number m.
so (k+1)k>m^2 (m^2-1) is the only solution here.
Which means k>=m^2, which means m-1
That trick at 2:26 is mind-blowing! 🤯
My method was to set n=2k (following form n(n+2)/4) becouse for this expression every even number will give an integer so then I got the equation k²+k=m⁴+m²-m+1 now move k²+k to rhs and factor in the following way (m²-k)(m²+k+1)=-(m-1) now becouse m,n≥0 it follow that k≥0 but by doing manipulations with that inequalities we get that m²-k≥0 and m²+k+1≥1 but so their product is at least 1 but becouse -(m-1) it would only be possible if m-1≤0 so m≤1 and we. have to check only m=0 and m=1 to get our pair (m,n)=(1,2)
PS:for the case when m²≤k it follows that m>1 but then we get (2m²)²
how is m^2-k >=0, what kind of manipulation should I use?
@@tianqilong8366 I should check 2 cases 1)when m²-k≥0
2)when m²-k≤0
Check my second comment in order to see the prove that there are no solutions for the second case
Nice and easy 👍
nice approach
Turn on postifications
that was such a beautiful solution mine was vastly more complicated 😭
Very easy sir.👍
OK,
Since RHS is integer, then n must be even. If we write n=2k, then
k*(k+1) = m^4+m^2-m+1
But a consecutive product is even, so, in the LHS, m must be odd (So no zero from the begining :))).
Now, the expresion m^4+m^2-m+1 must be a product of 2 consecutive numbers. In fact, we don't need n or k next, is anough just that LHS is a product of 2 consecutive numbers.
This product, obviously, will be "connected"with m^2-1, m^2, m^2+1.
So, we are going to prove that our expresion in m is bigger than a product of 2 consecutive numbers and less than the next product of 2 consecutive numbers.
First product: (m^2-1)*m^2
Let's prove (m^2-1)*m^2=0 which is allways true (discriminant is negative). More of that,
(m^2-1)*m^2 < m^4+m^2-m+1 (the inequality is strictly)
If we take the product of the other 2 consecutive numbers: m^2(m^2+1)
Let's prove m^4+m^2-m+1=1. But m is odd, so the inequality is allways true, with equal sign when m is 1.
So
(m^2-1)*m^2 < m^4+m^2-m+1 -m+1=0 => m=1
Therefore m=1 is the only solution. And then n=2, of course.
Quite simple with the inequality trick, but how did you think of that trick? Nice anyway 👍🏻
hes seen it many times before
I saw this trick some times to find when an expression is a square, it’s very useful to reduce the possible values of some numbers and then make calculuses with easy values. A simple examples are “find when 9n^2+36n+76 is a perfect square” or “solve 16n^2+7n+4=m^2”
It appears a lot in number theory problems. Especially when Michael Penn is solving them!
Whybeven add 1 after multiplying bith sides by 4..why not just set both sides as a product..n can equal 4 and then n plus 2 equals equals rest ofnthe expression and solve or n equals 2 and n plus 2 equals 2 times all the other terms and so on..
These problems where the only solution is the obvious one are getting a bit old.
nice solution