Could You Make The Greece IMO Team? | Greece IMO TST 2013

[tianqilong]

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<=0, m<=1,
so m can only be 0 or 1, then the rest is what the video has shown.

[brinzanalexandru]

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)

[bigalxyz]

Quite simple with the inequality trick, but how did you think of that trick? Nice anyway 👍🏻

[leif]

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..

[mathcanbeeasy]

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<=m^4+m^2-m+1 we get 2m^2-m+1>=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<=m^2*(m^2+1), which is equivalent to m>=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^2*(m^2+1)

So m^4+m^2-m+1 is bigger strictly than a product of 2 consecutive numbers, and is less or equal than the next product of the consecutive numbers.

So, m^4+m^2-m+1 can be only equal to m^2*(m^2+1).
m^4+m^2-m+1=m^4+m^2 => -m+1=0 => m=1
Therefore m=1 is the only solution. And then n=2, of course.

[brendanmiralles]

that was such a beautiful solution mine was vastly more complicated 😭

[willbishop]

These problems where the only solution is the obvious one are getting a bit old.