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$2\mid r^2+1$ forces $r$ odd. $pq = r+1$ even forces at least one of $p,q$ to be $2$. Plug in, and solve the easy two equations system to get $\{p,q\} = \{2,3\}$ and $r=5$.

$pq=r+1$ Thus $p$ or $q$ is even than we have $(p;q;r)=(2;3;5)$ or $(p;q;r)=(3;2;5)$

Just use some parity and get that one of $p$ or $q=2$. This immediately solves the problem. Was a very trivial one...

From $2(p^2+q^2)=r^2+1$ we see that $r$ is odd. Now from $pq = r+1$, if $r$ is odd clearly one of $p$ or $q$ is even. WLOG let $p=2$ since $2$ is the only even prime. Our equations are now $$2q=r+1$$and $$2(4+q^2)=r^2+1$$Letting $r=2q-1$ we see that after solving the quadratic equation that $q=3,-1$ and clearly $q$ cannot be negative. So $q=3$. It follows that our solutions are $(p,q,r)=(2,3,5), (3,2,5)$.

It was pretty trivial

$2(p^2+q^2)=r^2+1$ means $r$ is odd, so $r+1$ is even. Thus, $pq$ is even. But $p,q$ are primes, thus either of $p$ and $q$ is 2. WLOG, let $p=2$ $r=2q-1$, $r^2=2q^2+7$ Solving, we get $q=3, r=5$ Taking $q=2$ will give $p=3, r=5$ So, $(p,q,r)=(2,3,5),(3,2,5)$

Put $r=pq-1$ in equation 2. Then take mod P and Q. You get $p | 2(q^2-1)$. Similarly $q | 2(p^2-1)$. If $ p|2$,$ p=2$. Similarly if $ q|2$, $q=2$. But $p=q=2$ does not satisfy the equation with $r$ as a prime. Therefore wlog, $p=2$, $Q | (p^2-1)$, therefore $q=3$ and $r=5$. If $p|q-1$, $p+1 < q$. Therefore $q|p+1$ as if $q|p-1$, $p-1> q$ which is not true. Therefore $q=pk+1$ where k is a natural no. Therefore, $pk+1|p+1$ therefore $k=1$ $p+1=q$. We get $ p(p+1) -1=r$. Put those in eqn 2 and solve you will not get any natural no. as solution for p. Therefore $p,q={2, 3} or {3, 2}$ and $r=5$

I have a solution which uses mod 3 instead of mod 2. So here goes my solution : suppose all of p, q, r are not equal to 3. So this means that all of p^2, q^2, r^2 are congruent to 1 mod 3. So, 2(p^2 + q^2) is congruent to 1 mod 3, r^2 + 1 is congruent to 2 mod 3, which contradicts the fact that 2(p^2 + q^2) = r^2 + 1. Thus, at least one of p, q, r is 3. CASE 1: r=3 So this means that 2(p^2 + q^2) = r^2 + 1 = 3^2 + 1 = 10. But this is a contradiction as 2(p^2 + q^2) is at least 2(2^2 + 2^2) = 2(4 + 4) = 2(8) = 16. CASE 2: p=3 Thus we have 3q = r + 1, 18 + 2q^2 = r^2 + 1. The first equation gives q^2 = [(r+1)^2]/9 , the second equation gives q^2 = (r^2 - 17)/2. Thus we obtain [(r+1)^2]/9 = (r^2 - 17)/2 which just simplifies to (7r + 31) (r - 5) = 0. therefore, r=5. which gives q= (r+1)/3 = 2. Thus, (p,q,r) = (3,2,5) CASE 3: q=3. This case analogous to CASE 2 , and hence we obtain (p,q,r) = (2,3,5). Hence, all triples (p,q,r) are : (3,2,5) , (2,3,5). It is easy to check that all of these indeed satisfy the required conditions. th

From the second equation, it is clear that r is odd. Then, from the first equation, either p or q is even. Case 1: $$p=2$$.We get two equations in q and r. Solving,$$q=3$$and $$r=5$$. Case2: $$q=2$$. We again get two equations in p and r. Solving,$$p=3$$and $$r=5$$.

djmathman wrote: Find all triples $(p,q,r)$ of primes such that $pq=r+1$ and $2(p^2+q^2)=r^2+1$. Clearly as $2|r^2+1$ so $r$ must be odd then As $p*q=r+1=$ even $\implies p=2$ or $q=2$ we have $2(p^2+q^2)=r^2+1\implies r^2-2r-15=0\implies r=3$ hence $r=3$ then $(p, q)\in(2, 3)$

pq = r + 1 and 2(p² + q²) = r² + 1 where p,q and r are primes On observing, we can clearly see that there's a symmetry between p and q. Case 1:- p = q = r = odd i.e. p,q and r are odd primes. pq = r +1 pq is the product of two odd primes. So, it must be odd but the r + 1 is even. Case 1 is rejected Case 2:- r = 2 (even) and p = q = odd pq = 2 + 1 and, 2(p² + q²) = 2² + 1 = 5 (not a multiple of 2) Case 2 is rejected. Case 3:- r = odd and p or q = 2 (even) WLOG, let p = 2 then we've 2q = r + 1 or, r = 2q - 1 and 2(4 + q²) = r² + 1 On substituting r = 2q - 1, we get 8 + 2q² = (2q - 1)² + 1 q² - 2q - 3 = 0 (q - 3)(q + 1) = 0 So, q = 3 or -1(Rejected) Therefore, q = 3 pq = r + 1 r = pq - 1 = 6 - 1 = 5 So, (p,q,r) = (2,3,5) Now, WLOG let q = 2 then we've 2p = r + 1 or, r = 2p - 1 and 2(p² + 4) = r² + 1 On substituting r = 2p - 1, we get 2p² + 8 = (2p - 1)² + 1 p² - 2p - 3 = 0 (p - 3)(p + 1) = 0 So, p = 3 and -1(Rejected) Therefore, p = 3 Now, pq = r + 1 r = pq - 1 = 6 - 1 = 5 So, (p,q,r) = (3,2,5) Therefore, (p,q,r) = (2,3,5) and (3,2,5)

if p and q >2 then r is even and r=2 give contradiction.p and q symmetric in the equation.So case )p=2 and case)q=2 are similiar.Let p=2 we get q^2-2q-3=0-->(q-3)(q+1)=0 and q>0 --->q=3.2q=r+1 so r=5. Finally we have (p,q,r)--->(2,3,5),(3,2,5)