Helium wrote: Find all the pairs of integers $(m,n)$ satisfying the equality $3(m^2+n^2)-7(m+n)=-4$ Equation is $(6m-7)^2+(6n-7)^2=50$ Which easily gives $\boxed{(m,n)\in\{(1,0),(0,1),(2,2)\}}$

We see that: By factorization: $$(m-2)(3m-1)=(2-n)(3n-1)$$Then we examine the result: 1). When $m=n$, we have $m=n=2$. So, $(m,n)=(2,2),(\frac{1}{3},\frac{1}{3})$ is a solution. Also, we have, $$(6m-7)^2+(6n-7)^2=50$$So, we get the solution as $(m,n)=(0,1),(1,0)$. Hence, we have $$\boxed{(m,n)=(2,2),(0,1),(1,0)},\text{Beside that,},(\frac{1}{3},\frac{1}{3}) \text{is also a solution}$$And we are done! Thanks @Tuzo!

Wha do you think that $1/3$ is integer?

TuZo wrote: What do you think that $1/3$ is integer? Okay, sorry! I wanted to mention that aside but forgot!

How did you get $(6m-7)^2+(6n-7)^2=50$?

Helium wrote: How did you get $(6m-7)^2+(6n-7)^2=50$? Just expand them then you will find out. And, I arranged into that to get the result!

sorry but could you please expand that?! see I am just a beginner!

You want to expand ax^2+bx. In this problem, b=7 is an odd number, so multiply 4a on each side, we can expand it to (2ax+b)^2.

We write our equation again: $3m^2-7m+3n^2-7n+4=0$ $\Delta=49-36n^2+84n-48\ge 0$ $1+84n\ge 36n^2$ Here we get possible values of $n=0,1,2$ $n=0 \longrightarrow m=1$ $n=1 \longrightarrow m=0$ $n=2 \longrightarrow m=2$

Helium wrote: sorry but could you please expand that?! see I am just a beginner! $$36m^2-84m+49+36n^2-84n+49=50$$$$36(m^2+n^2)-84(m+n)+48=0$$Divide this expression by 12, then you will get your result!

https://artofproblemsolving.com/community/q3h1337550p7246328

using the fact $2(a^2+b^2) \geq (a+b)^2$ we get the relation $3x^2-14x+8 \leq 0$ where $x=a+b$ we have $\frac{2}{3} \leq x \leq 4$ hence the possible values are $0,1,2,3,4$ the rest is just cases

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