AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
is divisible by 
for all positive integers 
.
and satisfying 
.
be a polynomial in one variable with integer coefficients. Prove that the number of pairs 
of positive integers such that 
, is finite.
be a cyclic quadrilateral. Assume that the points 
are collinear in this order, in such a way that the line 
is tangent to the circle 
, and the line 
is tangent to the circle 
. Let 
and 
be the midpoints of segments 
and 
, respectively. Prove that the following three lines are concurrent: line 
, the tangent of circle 
at point 
, and the tangent to circle 
at point 
.
+1 w
be nonnegative real numbers such that 
. Let 
. Prove that 
and determine when the equality holds.
from the nonnegative integers to the integers satisfying 
and ![\[ f(mn) = f(m) + f(n) + f(m)f(n) \]](http://latex.artofproblemsolving.com/e/f/f/eff10a32cffaec742403ebfc46ff09bc8ef7b7c5.png)
for all nonnegative integers and .
, an arbitrary point 
is chosen on the side 
. In triangles 
and 
, the angle bisectors 
and 
are drawn, respectively. The point 
is the circumcenter of 
. Prove that the second intersection point of the circumcircles of triangles 
and 
lies on the line 
.
be a point in the plane of triangle 
such that the segments 
, 
, and 
are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to . Prove that 
is acute.
and be two distinct rays not lying on the same line, and let 
be a circle with center 
that is tangent to ray at 
and ray at 
. Let 
be a point on segment 
. The line through parallel to intersects line at . Let be the intersection of lines 
and , and let be the intersection of line and the line through parallel to . Prove that line 
is tangent to . be an acute triangle with point 
lie on the plane such that 
is a parallelogram. 
is the orthocenter of 
intersects at 
and 
intersects at 
The circle with diameter 
intersects the circumcircle of 
again at 
Prove that: The circumcircle of 
passes through the reflection point of over 
with 
, let 
denote the incenter and the midpoint of side . The line through perpendicular to 
intersects the tangent from to the incircle (different from line ) at a point > Show that is tangent to the circumcircle of triangle 
.
and 
. At each step, it is allowed to add to one of the polynomials and to multiply another one by the polynomial 
. Can the polynomials become equal at some point?
such that for any positive integer number 
and any arrangement of the numbers from 1 to 
on a circle, there exists a pair of adjacent numbers whose ratio (when dividing the larger number by the smaller one) is less than .
make the equation ![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
and 
satisfy the relations![\[ xy - zt = x + y = z + t. \]](http://latex.artofproblemsolving.com/c/1/2/c1270bde8804f79a317ab5f5ad6231b3fabcc4e4.png)
Is it possible that both 
and 
are perfect squares?
and 
and that 
so 
such that 
and 
then 
Claim
and 
then 
and 
which is a contradiction.
. Assume that 
or 
. WLOG 
, 
so we must have 
.
and 
so 
which is false. Hence our claim is proved.
and 
which can not hold. and can not be both perfect squares.
. Thus 
and 
for some 
. Similarly, take 
, 
and 
. Let 
, 
and 
. The equation is transformed to
where 
. First, note that 
thus 
. Similarly 
. As 
, we conclude that each prime divisor of 
is either 
or 
.
, 
and 
. Thus by 
, we also get 
and 
. As cannot divide both , we get 
.
must divide 
and cannot be congruent to 
, this number must be odd. This means all prime factor of must be . Thus
Now we will further factorize . Before that, note that imply 
where 
. Thus
By permuting if necessary, WLOG 
. Thus each prime factor of 
, 
, 
, 
is or .
is odd, 
is even., , , are odd. Thus they are all . Hence 
, contradiction as is even. and are odd. divides each of 
, , , and . Hence the denominator is divisible by 
. But the exponent of in the numerator is 
, contradiction.
cannot be a perfect square.
. WLOG 
and 
. Then
Since 
, it follows that 
are both even. Since 
, it follows that 
and 
for some positive integers . Thus 
. We set a system of equations
so
Then![$xyzt=\left[4a^2b^2-(a-b)^2\right]\left[4a^2b^2-(a+b)^2\right]=(4a^2b^2-a^2-b^2)^2-(2ab)^2$](http://latex.artofproblemsolving.com/6/4/3/6439d7f9e99d4f13b795a3b146857682ae91c5d8.png)
is a perfect square, it follows that
Thus 
. But then 
, contradiction.
exist then set 
and 
. Solving the equations we get:
we get:
This means they're product is a perfect square so:
By considering the minimum distance between adjacent squares this gives:
If 
we have:
is a perfect square which is a contradiction as 
. Therefore as 
(as 
positive) 
so we have:
Also:
Combining these results:
Hence 
or 
. As we've already shown 
(as positive) the only case left to consider if 
but:
giving a contradiction. and to both be perfect squares.
are of the form![\[\{\{x,y\},\{z,t\}\}=\{\{2cd-d+c,2cd-c+d\},\{2cd-c-d,2cd+c+d\}\}\]](http://latex.artofproblemsolving.com/2/d/7/2d78a3433cccccfa5d649d344407fc86ead1af58.png)
where 
are positive integers.
. Then, we have![\[x(k-x)-t(k-t)=k,\]](http://latex.artofproblemsolving.com/d/d/8/dd8e8a4c22a7c81eabc637d999a6dfbe7463bf28.png)
or![\[(x-t)(k-x-t)=k.\]](http://latex.artofproblemsolving.com/f/c/8/fc81ba4762b55c53fa77ab14b122e47442b23a76.png)
Without loss of generality, suppose 
(if 
, then 
so 
, which isn't possible). Let 
be a positive integer. We must now have that![\[x+t=k-k/a,\]](http://latex.artofproblemsolving.com/e/9/8/e987d767a914dcd96498edc2a2e6d75a09caafe0.png)
so 
, so let 
, where 
is a positive integer. Thus,
so
This implies that 
, or that 
, so both 
and are even. Let 
and 
where and are positive integers. We have that![\[x=2cd+c-d,\,\,\, t=2cd-c-d,\,\,\, k=4cd,\]](http://latex.artofproblemsolving.com/a/3/a/a3a1ead179373be66d6272f5daf41c0687b06b0c.png)
so the solutions are of the desired form. 
and were squares, and WLOG take 
. We see then that there exist positive integers and such that
or
Subtracting the equations gives![\[(a-b)(a+b)=4cd.\]](http://latex.artofproblemsolving.com/b/5/6/b56a54343fd04e27ea793f18571b5177dc66c124.png)
This implies that 
, and also that 
is even. If we had 
, then 
, so 
. Thus,![\[(2cd)^2-(c-d)^2\le(cd)^2,\]](http://latex.artofproblemsolving.com/c/b/9/cb9305661d0eec52f9eb1387219f7ca1dbe7aa45.png)
or 
. This is clearly impossible as 
and 
, so we must have 
.
, so 
and 
. Plugging this back in, we get that![\[c^2+d^2+1=3c^2d^2,\]](http://latex.artofproblemsolving.com/b/f/1/bf1183087f0f61c3924feef577af48674ad18353.png)
or![\[(3c^2-1)(3d^2-1)=10.\]](http://latex.artofproblemsolving.com/9/0/a/90a439dcb3439738b3801ea41b69b291155eb491.png)
It is easy to check that this has no solutions, so there are 
solutions.
and 
and
. Then 
If 
and , 
we get that
Now apply the four number theorem.
and 
which is true since 
and 
,
then
and 
and 
we must have 
, so 
and 
, impossible. and can't be both perfect squares .
such that 
, and define 
. Then, note that for any quadruple 
, we have 
, which is constant. Therefore, if we want the equality relation in the problem, we must have 
. As is unique, this means that for a pair of differences 
such that 
, there exists exactly one tuple 
which satisfies the problem assertion. Namely, 
both be squares. Then, we need pair such that 
, 
are both perfect squares. Letting 
, 
, we need 
to both be squares. First, note that if 
, then both of these numbers will be 
, so neither can be square. So, 
. Now, considering odd 
, where 
, we have that 
is a quadratic residue mod . As 
, we must have 
is a quadratic residue. Hence, 
. This argument can be repeated for all prime factors of 
, so 
. Now, 
, 
. So, the only way such that 
and 
can have the same 
is if 
is odd. To finish, note that 
, which is a contradiction.
.: 
is even, 
for some integer .
, WLOG assume 
non-negative integers(that is 
). Then, 
from the condition.
. So 
. are perfect squares, then 
and 
are squares.
is a perfect square, and 
is a perfect square.
.
. Obviously is even. Since are positive integers, so are 
( might be 
though). Note that 
.
then 
will be a contradiction as difference of the squares is too large so in this case 
.
or 
. So we have 
contradiction. as is even. Thus 
.
thus 
. is a positive integer, 
and so 
.
.
. The latter term is negative since is positive integer so it cannot be a square. But this contradicts both the facts that is positive and that squares are positive numbers. So, both cannot be simultaneously squares in this case.: is odd, 
.
assuming WLOG that are positive(
that is). Thus 
.
. Obviously is positive integer and 
.
are squares then 
and 
for some integers 
.
which is analogous to the equation we got earlier. Use the exact same logic as earlier to show that no solutions exist here too. Proved
, then 
and similarly, 
so 
, which is impossible since 
. So is even.
be such that 
We wish to show that they do not exist.
. Suppose otherwise, then, 
, but we clearly need 
, so 
which is impossible. Proven.
satisfy 
, then there exists integers 
such that either 
or 
Proof: Let 
be prime factors(possibly repeated) of . Note the following:
, then 
and 
are also sum of squares., there exist 
such that 
and 
, 
.
. It suffices to check for 
.
, suffices to check any 'combi' with the 3 others, details omitted) to 
: Since by removing 
, we can use the result for to write it as
We reduce it to the case.
Let 
. If 
we have either 
, which happens when 
or 
, the only possible case is when 
, and 
(or the constant 2 goes to the first one, which is the same thing WLOG). Implying that 
is a permutation of 
which is clearly impossible, as it renders at least 2 of 
having the same absolute value.
so we must have 
and 
which again gives solutions where 2 numbers have the same absolute value.
. If 
, then is odd and . Similarly, , a contradiction. Thus, 
.
and note
Shorthanding 
(because we know are the same parity), we wish to find positive integers , for which![\[4m^2n^2 - (m+n)^2, \quad 4m^2n^2 - (m-n)^2 \quad \text{are both squares}.\]](http://latex.artofproblemsolving.com/8/4/5/84513261d49e8bdbcf713f8c2bea2a86dea719ca.png)
Thus, their product is a square. However, note upon expansion the inequality![\[(4m^2n^2 - m^2 - n^2 - 1)^2 < (4m^2n^2 - (m+n)^2)(4m^2n^2 - (m-n)^2) < (4m^2n^2 - m^2 - n^2 )^2.\]](http://latex.artofproblemsolving.com/d/f/5/df52849088e133bb3aec24ff41089465e3f1b2e9.png)
Therefore, we have no solutions.
. We will consider a solution and determine what conditions it satisfies by solving the diophantine . WLOG assume 
and 
.
is not odd. is even, so is even. Similarly, is even. This implies that 
is even, a contradiction. 
. Let 
for integers 
with 
. We have![\[2d=xy-zt=d^2-k^2 - (d^2-\ell^2) = \ell^2-k^2.\]](http://latex.artofproblemsolving.com/6/7/2/6723693d943d9a7709be9edcb53c7ef653681eeb.png)
Observe that 
and are the same parity. Let 
. Note that both and are integers and 
. Clearly, we have 
because we would otherwise get 
. Thus, we have 
. Observe that![\[2d = (a+b)^2 - (a-b)^2 = 4ab,\]](http://latex.artofproblemsolving.com/a/d/a/ada413588feff280b3c34183a9670ac51e21a95c.png)
so we have the parametrization![\[(x,y,z,t) = (2ab+a-b,2ab-a+b,2ab+a+b,2ab-a-b).\]](http://latex.artofproblemsolving.com/0/d/3/0d3f780781777acacb8083e7d1e7117a0c79858a.png)
Note that 
does not work because it gives 
; thus, we have 
.![\[xy=4a^2b^2-a^2+2ab-b^2, zt = 4a^2b^2-a^2-2ab-b^2.\]](http://latex.artofproblemsolving.com/e/1/3/e131aee08d9752e8252535548e216e82eab58420.png)
Let 
. We have 
. It is clear that and have the same parity, so we get that 
, implying 
. Then, we get that![\[(ab-1)^2 \ge n = 4a^2b^2-(a+b)^2.\]](http://latex.artofproblemsolving.com/8/8/9/88932ad3610e668abc9764f7e39d1cb9086a5751.png)
Equivalently, we have![\[a^2+2ab+b^2 = (a+b)^2 \ge 4a^2b^2-(ab-1)^2 = 3a^2b^2 + 2ab - 1,\]](http://latex.artofproblemsolving.com/5/3/a/53a3058263911aad566155a24f5a1e6ec467866f.png)
so![\[a^2+b^2 \ge 3a^2b^2-1 > 2a^2b^2 = a^2b^2+a^2b^2 \ge a^2+b^2.\]](http://latex.artofproblemsolving.com/9/6/f/96fefc11d3a2ad542a4f7e1b2918408319fd55e0.png)
This is clearly absurd, so we are done.
given the equations.
. Then 
. We get the key factorization:![\[ a=x(a-x)-z(a-z)=(x-z)(-x-z+a)=(x-z)(y-z). \]](http://latex.artofproblemsolving.com/9/4/7/94771313ea45b25e0b52bf4ffded50f7e7153f32.png)
Let 
with 
. WLOG 
. Then 
. Now 
, so 
. Then 
, and 
, and 
. So we have the explicit parametrization![\[ (x,y,z,t) = \left(\frac{cd+c-d}{2}, \frac{cd-c+d}{2}, \frac{cd-c-d}{2}, \frac{cd+c+d}{2}\right). \]](http://latex.artofproblemsolving.com/2/e/5/2e5a5f19df0885aaa9b94eec4317beb728cf5c1d.png)
This proves the claim. 
and are squares. We have
for some 
.
.
. We claim 
and 
are of the same parity. This is clear from the definitions of and , since 
and 
are of the same parity. It suffices to show 
. Indeed,![\begin{align*}
(u-2)^2-v^2 &= \big[[(cd)^2-(c-d)^2]- 4\sqrt{(cd)^2-(c-d)^2} + 4\big] - \big[ (cd)^2-(c+d)^2\big] \\
&= 4cd - 4 \sqrt{(cd)^2-(c-d)^2} + 4 > 0.
\end{align*}](http://latex.artofproblemsolving.com/3/c/9/3c991fc6cf4b6ef37e47d6667f2a48d0fcab3f08.png)
This proves the claim. 
. Then 
.
which is clearly true. ![\begin{align*}
(u-4)^2-v^2 &= \big[[(cd)^2-(c-d)^2] - 8\sqrt{(cd)^2-(c-d)^2} + 16 \big] - \big[(cd)^2-(c+d)^2\big] \\
&= 4cd - 8\sqrt{(cd)^2-(c-d)^2} + 16 = 4cd - 8(cd - \epsilon) + 16 \\
&= -4cd + 8\epsilon + 16 < -4cd+8c+16=16+4c(2-d).
\end{align*}](http://latex.artofproblemsolving.com/e/6/1/e61fef2924ed57280df1e18115f01b70a72309ab.png)
Unless 
, we get a contradiction. Clearly 
; if 
or 
then 
. We need to check 
. It is not hard to check that none of these work, so we conclude that there are no solutions.
are all even. Suppose otherwise, and that 
are odd. Then, one of 
is even, and one of 
is even. Hence 
is even, a contradiction.
hence 
and 
for some positive integers 
. Plugging in, we get 
, and we want to determine if 
and 
can both be perfect squares. must be of the same parity, or else their difference can not be even. So let 
where is greater. Indeed, we get 
and we wish to determine if it is possible for 
and 
to both be perfect squares.![\[P = 16m^4n^4- 8(m^2 + n^2)m^2n^2 + m^4 - 2m^2n^2 + n^4.\]](http://latex.artofproblemsolving.com/1/b/e/1be1479fe126d89da41007b79baa1d3ff10aa094.png)
Clearly this expression is strictly less than 
since 
must be positive. Then, the stupid part is that somehow![\[(4m^2n^2 - m^2 - n^2 - 1)^2 < P\]](http://latex.artofproblemsolving.com/f/0/6/f0665f2b992bd5032bf246299268d1ad5ff9cff1.png)
holds since it simplifies to 
for all 
. A quick plug in shows that 
does not work. For all other cases, is bounded between two perfect squares, and hence cannot be a perfect square. So our answer is 
.
We can assume 
and since 
it follows that 
We have also 
so in fact 
Now write 
as 
From here we obtain (given that 
) 
thus 
(note that this is also true when 
)
Indeed
if this is not the case then WLOG is odd (the other case is the same) thus 
are odd and so is even We have that is also even therefore 
are even but in the end it doesn't even matter (
)
We surely cannot have 
and by the previous paragraph it follows that 
Now use 
so 
This is our favourite high-school quadratic inequality in 
so our first job is to calculate the roots of the equation 
namely 
We easily check that 
so it must be the case 
We are ready for the contradiction The final inequality is
as desired 
then exist prime number 
that divides . is prime number and divides 
for some integer then 
or 
for some integer . are all odd.
but 
. Contradiction!
is even. We have 
now if 4 divides then with Lemma-1 there exist prime number that divides 
and so divides but with Lemma-2 this is Contradiction! It means 
but we know is perfect square and it means 
is even number so we have 
too. 
and 
are even numbers so is even and with above proof we conclude 
but in this case we have 
. If , then is odd and . Similarly, , a contradiction. Thus, .Now note and note
(because we know are the same parity), we require to find to find positive integers , for which
Thus, their product is a square. However, note upon expansion the inequality 
< 
Therefore, we have no solutions.
and 
are distinct modulo , so and are both even and so is their difference which gives a contradiction. 
. Let and . Define such that :
and 
are of the same parity, so their differences must be atleast . This gives 
and 
. Squaring both sides gives 
. Similarly 
.
then exist prime number that divides . is prime number and divides for some integer then or for some integer . are all odd. but . Contradiction! is even. We have now if 4 divides then with Lemma-1 there exist prime number that divides and so divides but with Lemma-2 this is Contradiction! It means but we know is perfect square and it means is even number so we have too. and are even numbers so is even and with above proof we conclude but in this case we have requires more explanation.
, hence 
, contradiction. 
and 
, then 
.
, then 
. Fixing 
, we have
Let 
, then 
, hence
Therefore, 
, hence is even, let 
, then 
. Let 
, then
Therefore, we conclude that
Now it is just some simple bounding, notice that 
, hence 
. Notice that
If 
then we need 
, no solution.
, we need 
, no solution.
then 
hence 
, which gives no solution.
.
and 
. quadratic equations:
Let 
and 
be the discriminants of the given quadratic equations, respectively.
Since we have that 
, we have that 
.
.
, then there exist such that 
and all of their respected permutations.
and 
, 
and 
.
, and having that , we get that:
and are both square numbers.
and 
are also square numbers.
thus we can claim wlog that 
and 
.
, we get a contradiction, since any solution we got, brought us that and aren't squares at the same time.
, then 
. Note that 
, so 
![\[(x-y)^2=(x+y)^2-4xy = (a^2-b^2)^2-4a^2\]](http://latex.artofproblemsolving.com/3/8/2/382ac17f00a26ace1d32f7ec4838230e4132846e.png)
![\[(z-t)^2=(z+t)^2-4zt = (a^2-b^2)^2-4b^2\]](http://latex.artofproblemsolving.com/b/2/a/b2ae34db7c0067a261ea0fd03173d29d24c9223e.png)
Note that must be of the same parity mod 2. AFTSOC otherwise, then is odd, so both must be even, contradiction.
, where 
. Then, we now know that 
is a square and equals:![\[((a^2-b^2)^2-4a^2)((a^2-b^2)^2-4b^2)=(16m^2n^2-4(m+n)^2)(16m^2n^2-4(m+n)^2)\]](http://latex.artofproblemsolving.com/b/c/9/bc914dae466a04ca34afb6b4b035685da1a1a212.png)
Which equals,![\[=16(4m^2n^2-m^2-n^2-2mn)(4m^2n^2-m^2-n^2-2mn)\]](http://latex.artofproblemsolving.com/a/f/b/afba2f1cdf621bc89fe3a82a7586790a75883979.png)
We can pull out 
, so,![\[S= (4m^2n^2-m^2-n^2)^2-(2mn)^2 \text{ is a square}\]](http://latex.artofproblemsolving.com/7/b/d/7bdaa7124e40a7fcd2c6ea84ecc01b62d811e82a.png)
Note that for .![\[ 0<(2m^2-1)(2n^2-1)-2 \Longrightarrow 4m^2n^2 < 8m^2n^2-2m^2-2n^2-1\]](http://latex.artofproblemsolving.com/4/6/b/46b05408954035cad1795f7460771fec11cc65c8.png)
Thus, 
can be bounded between two consecutive squares![\[(4m^2n^2-m^2-n^2-1)^2< (4m^2n^2-m^2-n^2)^2-(2mn)^2 < (4m^2n^2-m^2-n^2)^2\]](http://latex.artofproblemsolving.com/e/a/3/ea38f3cb0728dd88227286d65b279af70b8c87c8.png)
thus, we have a contradiction, so cannot ever both be squares..
with in that order, and let 
, where WLOG 
so 
(else 
). Then we have 
and 
, so
so we want 
and 
to be perfect squares. Evidently, and are the same parity, otherwise 
is even but 
is odd so we have 
. As such, we can let 
, so 
and we have 
. Then
so 
. Then,
and
where 
are nonnegative integers. Then
where the last line comes from the fact that 
makes the LHS zero, while 
clearly fails as 
. This means that 
, so
For a fixed value of 
, the RHS is minimized when one of 
is equal to , so setting , we need 
, which only holds for 
. Then 
, but this implies that 
—impossible. As such, no such positive integers 
exist. 
After division on all possible common divisors we can go to the problem:
for positive integers 
with 
, where 
.
.
For every prime , dividing , it is also true that
from 
and 
from 
.
, otherwise 
would be odd, so should divide but not 
- contradiction with .
and there are coprime positive integers 
such that
therefore it follows from 
that
Analogously,
and we get from 
that
Adding 
to 
, we find that
therefore the next divisibility is true from 
Let us remember that is odd, then should be even, so divisibility below is also true
It is easy to see that 
is true iff
, that contradicts to 
, q.e.d.
, 
for some integer . Then 
. Let 
, then 
and 
. It follows that 
, which is a perfect square. Similarly, 
, ![\[y-x=z-a-(t+a)=z-t-2a=b-a \implies (ab)^2-(a-b)^2=4xy\]](http://latex.artofproblemsolving.com/b/2/0/b20293ef3c740310fecdeee53670d9f1e92f612a.png)
another perfect square.
and 
. Then ![\[(ab-k+l)^2-(ab-k)^2=4ab \implies 2l(ab-k)+l^2 = 4ab\]](http://latex.artofproblemsolving.com/b/3/7/b374b4b7834143cbaae873039c2f5cc2dae38bc4.png)
![\[(ab)^2-(a+b)^2 < (\frac{1}{2}ab)^2 \implies \frac{3}{4}ab^2 < (a+b)^2 \implies a+b > \frac{\sqrt{3}}{2}ab\]](http://latex.artofproblemsolving.com/4/8/3/4834d6ce372d28e0ae8f7bb01921c7b333cda56a.png)
Note that 
, else would be nonpositive. In particular, at least one of must be at least 
, and both must be at least . Thus, WLOG let 
. But dividing by we have ![\[1+\frac{b}{a} > \frac{\sqrt{3}}{2}b \implies 1+\frac{b}{3} > \frac{\sqrt{3}}{2}b \implies \left( \frac{\sqrt{3}}{2}-\frac{1}{3} \right) b -1 < 0 \]](http://latex.artofproblemsolving.com/d/9/8/d987af0af8ad8d686c3bb7554df6497fb005455a.png)
But after rearranging it is apparent that this implies 
, contradiction.
. Furthermore, 
is even for parity reasons. Thus 
. Hence 
. But then 
,contradiction and we are done.
Then cannot be smaller than both and , as this would mean 
impossible. So assume 
as well.
If 
then 
absurd. Thus 
so 
This gives the bound 
Now assume and are both perfect squares. Then there exists 
such that 
Rearranging, 
Hence or 
so we have reduced the problem to a finite case check.
.
such 
.
and 
. to be even, so 
must both be even, and thus 
. to be a multiple of , then let 
, so 
Then, 
and 
are perfect squares, and their difference 
. As such, by difference of squares, 
and 
. Similarily, 
holds when not both are .
, contradiction.
cannot be a perfect square.
; WLOG 
by parity arguements, let
,notice that must divide both of these. Let 
, 
. It follows that 
and 
now to show that this is not a perfect square. We use bounding,
so there exist positive integers such that 
Then from we get 
which rearranges to 
Then notice that are all odd, so 
are even so by symmetry we can assume they are both 
as opposed to 
Then letting 
we get 
Then we have 
and 
First, we notice that is invalid and for all other pairs of positive integers we have 
so 
Thus if we let 
and 
for positive integers 
we have that 
and 
Thus if 
we see that 
but 
so 
Now if 
we see that 
so 
so 
so we are done.
and . Clearly, 
but which means that 
. WLOG, assume 
. Then,![\[(a-b)(a+b)=a^2-b^2=xy-zt=xy-z(x+y-z)=(z-x)(z-y)\]](http://latex.artofproblemsolving.com/8/e/d/8ed12b8a911b614f41c4a37f25d176ca64a16b2d.png)
so by the four-number theorem we can write 
, 
, 
, and 
for positive integers . Then we can easily find that 
so
Then using 
we can get 
. We have either 
or 
, which implies 
. WLOG, assume 
then![\[s^2q^2r^2=(1+s^2)(q^2+r^2)\le 2s^2(q^2+r^2)\]](http://latex.artofproblemsolving.com/a/4/5/a45305cf13c5842cdce6940a9f0d59dbb4258628.png)
so 
and it is clear at this point that 
and 
which gives us the values 
and 
which is absurd.
; let 
, 
, 
, 
. Without loss of generality, 
. Suppose to the contrary that and are perfect squares; write 
, 
. Then
Additionally, we have
Lemma 1. 
.
for 
; then since 
, we have 
for 
, similarly 
for 
. Then note that
and the corresponding equation for 
, from which it is evident that 
is even, a contradiction. 
and 
it follows that 
, for 
. Then since 
, we get 
and 
and similarly 
, 
for 
. Thus (1) and (2) become
Lemma 2. For positive integers 
, if we have 
then 
.
, so 
. Then 
, so we must have 
. 
, etc. and substituting into (
) and (
) we get
So we can apply the lemma again to get 
. Continuing this process yields 
for any 
, which is obviously impossible. But since 
satisfy the conditions of the lemma, we have a contradiction, so we are done.
and 
. Relabel so that is now . Then our condition becomes ![\[p^2-q^2=\frac{p^2+x^2}{x}=\frac{q^2+y^2}{y}\]](http://latex.artofproblemsolving.com/0/d/9/0d9e8c8b9fac9fb303bbfb2491a80ff8d0043f35.png)
Turning these into quadratic equations we get
and 
need to be perfect squares. Now, note that if 
Then, letting 
, we notice that ![\[(x+4)^2-x^2 > \left(\frac{p^2-q^2}{2}+2\right)^2 - \left(\frac{p^2-q^2}{2}-2\right)^2 = 4(p^2-q^2) \implies (x+4)^2 > (p^2-q^2)-4q^2\]](http://latex.artofproblemsolving.com/2/b/8/2b8e03bc52f48cb852fc2ea4e91f2b04c14587c3.png)
Hence, 
, by parity. This implies 
and hence 
. This obviously cannot be true since it would imply that 
is not congruent to modulo .
. Then, we would have 
which implies that does not exist, a contradiction. 
bound on 
,
bound; this can be seen by plugging in small values of 
is odd, 
contradiction. Thus we assume that 
This now transforms the equation into
and we want to find out if both 
and the latter as 
.
,
is odd, thus giving us a contradiction. So 
is at least 
.
.
,
and various permutations. This is a contradiction as this does not give four positive 
,
We are given 
We have 
and 
are both perfect squares. Then take 
so then 
But 
is not a perfect square because it can be bounded between 
and 
.
.
and 
.
is equal to 
,
.
.
and 
.
.
,
,