WakeUp wrote: Find all functions $f:\mathbb{Z}_{>0}\rightarrow\mathbb{Z}_{>0}$ that satisfy the following two conditions: $\bullet\ f(n)$ is a perfect square for all $n\in\mathbb{Z}_{>0}$ $\bullet\ f(m+n)=f(m)+f(n)+2mn$ for all $m,n\in\mathbb{Z}_{>0}$. Second line shows that $f(x)$ is strictly increasing. Let $f(x)=g(x)^2$ with $g(x)>0$ strictly increasing too. So $g(m+1)\ge g(m)+1$ So $g(m+1)^2=g(m)^2+g(1)^2+2m\ge g(m)^2+2g(m)+1$ and so $g(m)\le m+\frac{g(1)^2-1}2$ So $g(m+1)>g(m)+1$ can occur only a finite number of times and so $g(m)=m+c$ $\forall m$ great enough. Pluging this in original equation, we get $c=0$ and so $g(m)=m$ $\forall m$ great enough and so $g(m)=m$ $\forall m$ since positive and strictly increasing. Hence the unique answer $\boxed{f(n)=n^2}$ which indeed is a solution

$f(n+1)-f(n)=2n+f(1)$ And hence, $f(n)=n(n+c) ; c=f(1)-1$ Since $4f(n)=(2n+c)^2-c^2$ and RHS cannot be a square for large $n$ unless $c$ is zero, we conclude $c=0$ and hence the only solution $f(n)=n^2$

$f(2)=f(1)+f(1)+2=2f(1)+2$ $f(3)=f(2)+f(1)+4=3f(1)+6$ $f(4)=f(3)+f(1)+6=4(f(1)+3)$ Since $f(4)$ is square, so is $4(f(1)+3)$ which means $f(1)+3$ is a square. But $f(1)$ is a square also, and the only squares that differ by $3$ are $1^2$ and $2^2$ so that $f(1)=1$. To prove that $f(n)=n^2$ for all $n$, the base case is done, and for the inductive step note that $f(m+1)=f(m)+f(1)+2m=m^2+2m+1=(m+1)^2$. So $f(n)=n^2$ for all $n$. lajanugen's solution doesn't make any sense.

Maybe a closer look at lajanugen's solution will help. Let us take it step-by-step. For $m=1$ we get $f(n+1)-f(n)=2n+f(1)$, for all $n\geq 1$. This is a linear recurrence relation for the sequence $a_{n+1} = a_n + (2n+f(1))$, by taking $a_n = f(n)$. Denote $b_n = n^2 +n(f(1)-1)$; it is immediately checked that we also have $b_{n+1} = b_n + (2n + f(1))$. Subtracting the two relations yields $a_{n+1} - b_{n+1} = a_{n} - b_{n}$, thus constantly equal to $a_1 - b_1 = 0$. And hence, $f(n)=n(n+c)$, for $c=f(1)-1$. Since then $4f(n)=(2n+c)^2-c^2$ (quite easy to check), and $f(n)$ is given to be a perfect square $x_n^2$, it follows $(2n+c)^2 - x_n^2 = c^2$ for all $n\geq 1$, impossible if $c\neq 0$, since the difference between two unequal squares indefinitely increases when one of them indefinitely increases. We conclude $c=0$ and hence the only solution $f(n)=n^2$. It remains to be seen that this indeed verifies the conditions, which it does. So I suggest, dear WakeUp, to wake up and revise your rating of lajanugen's post!

OK! I've woken up! I never meant that it was incorrect, but that it was extraordinarily vague.

thankyou very much mavropnevma, for taking your time to comment on my solution

I did not read the solutions. So I don't know whether this logic has been posted before It is easy to see that f(n)=n(n-1+f(1)) Now choose n to be a prime p Clearly p divides f(1)-1 For any prime this holds So a finite number is divisible by infinitely many primes This implies f(1)=1

let $f(1)=c^2$,then$f(m+1)=f(m)+2m+c^2$ hence $f(m)=f(1)+(m-1)c^2+2[1+...+(m-1)]=m(c^2+m-1)$ if $c>1$ let m be a prime greater then $c^2-1$,then $p||f(m)$ contradiction! hence $c=1,f(m)=m^2$.

Defining a new function $g(x)=f(x)-x^2$, we note that $g(m+n)=g(m)+g(n)$.However, this is the Cauchy equation, whose solutions we know to be $g(x)=cx$.This means that $f(x)=x^2+cx$.Substituting $x=c$, we see that $f(c)=2c^2$.However,$2c^2$ is not a perfect square unless $c=0$.Thus,$f(x)=x^2$,which works.

WakeUp wrote: Find all functions $f:\mathbb{Z}_{>0}\rightarrow\mathbb{Z}_{>0}$ that satisfy the following two conditions: $\bullet\ f(n)$ is a perfect square for all $n\in\mathbb{Z}_{>0}$ $\bullet\ f(m+n)=f(m)+f(n)+2mn$ for all $m,n\in\mathbb{Z}_{>0}$. Almost the same solution as @above so i just post this for storage. Let $g(n)=n^2-f(n)$ Multiply both sides of the equation by $-1$ to get: $$-f(m+n)=-f(m)-f(n)-2mn$$And now add $(m+n)^2$ on the both sides of the equation to get: $$g(m+n)=(m+n)^2-f(m)-f(n)-2mn=m^2-f(m)+n^2-f(n)=g(m)+g(n) \implies g(n)=c \cdot n$$Plug it back on $g(n)=n^2-f(n)$ to get that $f(n)=n^2-c \cdot n$ and when you plug $n=c$ you get $c=0$ since $2c^2$ is an square if and only if $c=0$. Thus the only solution is: $\boxed{f(x)=x^2 \; \forall x \in \mathbb Z_{>0}}$ Thus we are done

Let $f(1) = c^2$ for some $c \in \mathbb{Z^+}$. Then by induction we have $f(n) = nc^2 + n(n-1) = n(c^2 + n - 1)$ for all $n\in \mathbb{Z^+}$. If $c > 1$, then choose a large prime $p$ such that $\gcd{(p , c^2 - 1)} = 1$. But then plugging $n = p$, we get that $p(c^2 + p - 1)$ is a perfect square. And by our choice of $p$, we have $\gcd{(p, c^2 + p - 1)} = \gcd{(p, c^2 - 1)} = 1$, so both $p$ and $c^2 + p - 1$ are perfect squares, contradicting the fact that $p$ is prime. Thus, $c = 1$ and we have our only solution $f(n) = n^2 \ \forall n \in \mathbb{Z^+}$ which clearly works. $\blacksquare$

Plug $n\mapsto 1$ in the second equation to get $f(n+1)-f(n)=2n+f(1)$. By simple induction, we have $f(n)=n(n-1+f(1))$ for all positive integers $n$. Let's consider prime $p$. We have $f(p)=p(p-1+f(1))$. Since $p\mid f(p)$ and $f(p)$ is a perfect square, we must also have $p^2\mid f(p)$ so $p\mid p-1+f(1)$. Thus, $p\mid f(1)-1$ for all prime $p$. Taking $p$ sufficiently large yield $f(1)=1$ so $f(n)=n^2$ for all positive integers $n$ which obviously works, as desired.

let f(x)=(g(x))^2 and g has the same domain as f, which is the natural numbers. Since f(x) is a perfect square, it is monotone increasing in the interval [1, ∞) and the naturals are a subset of that hence it is monotone increasing over naturals too. now we put m=n=k in the equation (g(m+n))^2=(g(m))^2+(g(n))^2 + 2mn , hence we get (g(2k))^2= 2((g(k))^2+k^2). Now since LHS is a perfect square and since it is even, it should be divisible by 4. So 2 | g(k)^2+k^2. This is only possible when g(k)^2= (2l+1)k^2 for some whole number l. However we need 2l+1 t be a perfect square for all l. hence l has to be zero. hence f(x)=g(x)^2=x^2. can someone check this solution, I haven't done much proof writing before, so is this fine?( apologies for not knowing latex)