it can be deduce quickly from Four Number Theorem

What is Four Number Theorem? PS. Ehsan! Please, read http://www.artofproblemsolving.com/viewtopic.php?p=140098 and name problems properly next time! Regards, Myth

Myth wrote: What is Four Number Theorem? The equation $xy=zt$ has only postive integers in forms $x=am,y=bn,z=an,t=bm$ where $(m,n)=1$

That is what I thought.

How do you use this Theorem?

use some facts like b+c\mid a^2 + bc b+c\mid c^2 + ad b+c\mid d^2 +bc b+c\mid b^2+ ad (for example, ad=bc then a^2 +ad =a^2+bc so we'll have a(a+d)=a^2+bc and because b+c|a+d we'll get b+c| a^2 +bc) after that try to use b+c|(b-a)(b+a) and b+c|(c-a)(c+a), btw besides b+c can not devide both c-a and b-a (unless a very special case as m<2) By these all you can get (a,b,c,d)=(1,2^(s)-1,2^(s)+1,2^(2s)-1)

Hint : 1/2[b+c]|c+a hence b+2a =<c 1/2[b+c] | b+a hence c=<b+2a so c= b+2a and :ad = b^2 +2ab, a^2+ ad= (a+b)^2 , a(a+d)=(a+b)^2 because b+c=2(b+a) so the only prime factor of b+a is 2,so a is a power of 2,and it's odd too,so a=1

Let $f:[1,b]\rightarrow \mathbb{R},\ f(x)=x+\dfrac{bc}{x}$. As $f^\prime (x)=1-\dfrac{bc}{x^2}\le 0$, we infer that $f(x)\ge f(b)=b+c,\ \forall x\in [1,b]$; in particular, $a+d=f(a)\ge b+c\Leftrightarrow k\ge m$. Now, $ad=bc\Leftrightarrow a(2^k-a)=b(2^m-b)\Leftrightarrow (b-a)(a+b)=2^m(b-2^{k-m}a)$, hence $2^m|(b-a)(a+b)$. It is easy to see that for $x,y\in \mathbb{Z}$, if $v_2(x\pm y)\ge 2$, then $v_2(x\mp y)=1$. If $v_2(b-a)\ge 2$, $v_2(a+b)=1$, so $v_2(b-a)\ge m-1\Rightarrow b>2^{m-1}$, which is in contradiction with the fact that $b<\dfrac{b+c}{2}=2^{m-1}$. Thereby, $v_2(a+b)\ge m-1,\ v_2(b-a)=1$. Write $a+b=2^{m-1}\alpha$. If $\alpha \ge 2\Rightarrow2^m\le a+b<b+c=2^m$, contradiction; so $a+b=2^{m-1}$ and $b-a=2\beta$, or equivalently $a=2^{m-2}-\beta,\ b=2^{m-2}+\beta$

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Substituting back, $(b-a)(a+b)=2^m(b-2^{k-m}a)\Leftrightarrow 2^m\beta=2^m(2^{m-2}+\beta-2^{k-m}a)\Leftrightarrow 2^{k-m}a=2^{m-2}$ As $m>2$ and $a$ is odd, we get that $a=1$. Furthermore, $k=2m-2$, hence $d=2^{2m-2}-1$. Now $b(2^m-b)=2^{2m-2}-1\Leftrightarrow \left ( b-(2^{m-1}-1) \right ) \left ( b-(2^{m-1}+1)\right )=0$ which together with the fact that $b<c$, we get that $b=2^{m-1}-1,\ c=2^{m-1}+1$, so the family of the solutions $(a,b,c,d)$ is described by the set $$ M=\{ \left (1,2^{m-1}-1,2^{m-1}+1,2^{2m-2}-1 \right )|\ m\in \mathbb{N},m\ge 3 \}$$

Note that, if $k\leqslant m$, then using $a+d\leqslant b+c$, and $ad=bc$, we get $(d-a)^2\leqslant (c-b)^2$, which is clearly impossible. Thus, $k>m$ must hold. Next, observe that, $b+c =2^m$ and $b<c$ implies $b<2^{m-1}$. Keeping this in mind, we now observe that, $bc=b(2^m-b)=ad=a(2^k-a)$, implies $b\cdot 2^m - a\cdot 2^k = (b-a)(b+a)$. Thus, $2^m \mid (b-a)(b+a)$. Now, note also that, if $d={\rm gcd}(b-a,b+a)$, then $d\mid 2b$, and thus, the largest power of $2$ dividing either $b-a$ or $b+a$ is exactly $2$. Clearly, $b-a<2^{m-1}$, and thus, $2^{m-1}\mid b+a$. Moreover, if $b+a\geqslant 2\cdot 2^{m-1}$, then we again have a contradiction, as $b,a<2^{m-1}$. Thus, $b+a=2^{m-1}$. This yields, $b-a = 2(b-a\cdot 2^{k-m})$, which brings us, $b=(2^{k-m+1}-1)a$. Now, using $b+a=2^{k-m+1}a = 2^{m-1}$, we immediately obtain $a=2^{2m-k-2}$. This immediately establishes (since $a$ is odd), that $a=1$.

$$ad=bc \implies a((a+b)-(b+c))=(a-b)(a-c) >0.$$It follows that $a+d > b+c, 2^k >2^m$ and $k>m.$ Since $ad=a(2^k-a)=bc=b(2^m-b$ we get $$2^mb-2^ka= b^2-a^2=(a+b)(a-b).$$Now $2^m(b-2^{k-m}a)=(b-a)(b+a) \implies 2^m|(b-a)(b+a).$ But since the difference between $b-a$ and $b+a$ is an odd multiple of 2, we have that 4 must not divide either one of them. It follows that $2^{m-1}|b-a$ or $2^{m-1}|b+a.$ But $0<b-a<b<2^{m-1} \implies 2^{m-1}|b-a.$ Now $0<b+a<b+c=2^m \implies b+a=2^{m-1} \implies c=2^{m-1}.$ So $$ad=bc=(2^{m-1}-a)(2^{m-1}+a) \implies a(a+d)=2^{2m-2} \implies a=1.$$Q.E.D.

dyta wrote: How do you use this Theorem? Let $a=xy, b=xt, c=yz, d=zt$ with odd positive integers $x,y,z,t$ such that $x<z$ and $y<t$. It is easy to see that $k>m>2$. Further, $$2^m(2^{k-m}+1)=a+b+c+d=(z+x)(t+y),$$$$2^m(2^{k-m}-1)=a-b-c+d=(z-x)(t-y).$$Of course, numbers $x\pm z, y\pm t$ are even. If both $x+z$ and $y+t$ are divisible by $4$, then $x-z, y-z$ are not divisible by $4$, so $m=2$, that's impossible. Similar things happen when both $x-z$ and $y-t$ are divisible by $4$, so we have two cases: 1) $\nu_2(z+x)=1, \nu_2(t+y)=m-1, \nu_2(z-x)=m-1, \nu_2(t-y)=1$. So $2(z-x)\geq 2^m=xt+yz$, hence $y=1$. Further, $2(t+1)\geq 2^m=xt+z$, therefore $x=1$, and we finally get $a=xy=1$. 2) $\nu_2(z+x)=m-1, \nu_2(t+y)=1, \nu_2(z-x)=1, \nu_2(t-y)=m-1$. Here $2(t-y)\geq 2^m=xt+yz$, hence $x=1$. Also we have that $2(z+1)\geq 2^m=t+yz$, therefore $y=1$ and $a=xy=1$ again. Let me underly an evident inflation of IMO problems, because similar problem P5 from IMO2015 was harder than this one.

Four number theorem does not exist. We claim that the solutions are of the form $(a,b,c,d) = (1,2^{m-1}-1, 2^{m-1}+1, 2^{2m-2}-1)$ and thus $k=2m-2$. Let \begin{align} a&=2^{k-1} - x\\ d&= 2^{k-1}+x\\ b &=2^{m-1} - y\\ c &= 2^{m-1}+y \end{align}where $2^{k-1}-1 \geq x>y\geq 1$ are odd integers. Then, the $ad=bc$ condition becomes \[2^{2k-2} - x^2 = 2^{2m-2} - y^2 \Longrightarrow 2^{2k-2} - 2^{2m-2} = x^2-y^2\] $\textbf{Claim: }$$k\geq 2m-2$. $\textbf{Proof: }$ Note that \[k\geq v_2(x^2-y^2) = v_2(2^{2k-2} - 2^{2m-2}) = 2m-2\]where the first inequality is true because $x,y$ odd implies that$\min \{v_2(x-y),v_2(x+y)\} = 1$, and $y<x<2^{k-1}$ implies that $\max \{v_2(x-y),v_2(x+y)\}\leq k-1$. $\square$ $\textbf{Claim: }$ $k\leq 2m-2$ with equality when $x=2^{k-1}-1$. $\textbf{Proof:}$ Note that \[2^{2k-2} - 2^{2m-2} = x^2-y^2 \leq (2^{k-1}-1)^2 - 1^2,\]thus $2^{2k-2} - 2^{2m-2} \leq 2^{2k-2} - 2^k$, so $k\leq 2m-2$. $\square$ Thus, we must have $k=2m-2$, and so the $x=2^{k-1}-1 \Longrightarrow a=1$ equality case must hold and we're done. $\blacksquare$.

Neither 4-Number Lemma nor LTE : Since $ad=bc$, we have: $$a((a+d)-(b+c)) = a^2+ad-ab-ac = a^2-ab-ac+bc=(a-b)(a-c)>0$$Therefore $a+d > b+c$ or $k>m$. Note that: $$ad=bc \implies a (2^k-a) = b(2^m-b) \implies (b-a)(b+a) = b^2-a^2=2^mb-2^ka = 2^m(b-2^{k-m}a)$$Note that $a, b, c, d$ are odd integers. Since $b-a, b+a$ have a difference of $2a$ which is an odd multiple of $2$, we must have either: $2^{m-1}|b-a$ or $2^{m-1}|a+b$. Notice that: $b-a < b < 2^{m-1}$ which implies $2^{m-1}|a+b$. Notice that: $a+b < b+c = 2^m$ which implies $a+b=2^{m-1}$. Also, we have: $b=2^{m-1}-a, c=2^{m-1}+a$. Thus: $$ad=bc=(2^{m-1}-a)(2^{m-1}+a) = 2^{2m-2}-a^2 \implies a(a+d)=2^{2m-2}$$Since $a$ is odd integer, $a=1$ and we are done.

Note that $2^{m - 1} < c < d < 2^k \implies m - 1 < k \implies m \leq k$. So, $2^m \mid a + d \implies ad \equiv -a^2\mod{2^m}$. Note that $2^m \mid b + c \implies bc \equiv -b^2 \mod{2^m}$, and $ad = bc \implies -b^2 \equiv -a^2 \mod{2^m} \implies 2^m \mid (b - a)(b + a)$. If $4 \mid b - a$, $b + a \equiv 2\mod{4} \implies 2^{m - 1} \mid b - a$, a contradiction as $2b < 2^m \implies b < 2^{m - 1}$. So, $4 \mid b + a$ and $b - a \equiv 2\mod{4} \implies 2^{m - 1} \mid b + a$. Since $b < 2^{m - 1}$, $a < 2^{m - 1}$ and $b + a < 2\cdot2^{m - 1} \implies b + a = 2^{m - 1}$. Hence, $2b + 2a = 2^m = b + c \implies 2a + b = c$, and $\gcd(a, b) = 1$. If $a \neq 1$, consider a prime $p$ such that $p \mid a$. $ad = bc \implies p \mid bc \implies p \mid c \implies p \mid 2a + b \implies p \mid b$, a contradiction, so $a = 1$.